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Basic Numeracy

295 questions

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Question 1Question →

In a sequence of arithmetic operations, a candidate calculates a value of −82-82. If this integer is then divided by 99, determine the least non-negative remainder.

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Answer: 8

Answer

8
Applying the division algorithm, −82-82 can be expressed in the form dq+rdq + r as 9×(−10)+89 \times (-10) + 8. Since the remainder rr must be greater than or equal to 00 and strictly less than the divisor 99, the least non-negative remainder is 88.

Step-by-Step Solution

1
Identify the dividend and the divisor.
Dividend is −82-82, divisor is 99.
To apply the formal mathematical definition of division with remainders.
2
Find the largest integer multiple of 99 that is less than or equal to −82-82.
The correct multiple is −90-90 (since −90≤−82-90 \leq -82).
The remainder must be a strictly non-negative value, meaning the multiple we subtract from the dividend must not exceed the dividend.
3
Calculate the difference between the dividend and this multiple to find the remainder.
−82−(−90)=8-82 - (-90) = 8.
This difference represents the least non-negative remainder required by the problem.

Key Concept

Remainder Theorem and Negative Dividends
Question 2Question →

Determine the positive remainder when the value of 364336^{43} is divided by 3737.

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Answer: 36

Answer

36
By the properties of modular arithmetic, 3636 is congruent to −1-1 modulo 3737. When we raise both sides to the power of 4343, we obtain 3643≡(−1)43(mod37)36^{43} \equiv (-1)^{43} \pmod{37}. Since an odd power of −1-1 is still −1-1, the expression leaves a remainder of −1-1. To find the standard positive remainder, we simply add the divisor to this result: −1+37=36-1 + 37 = 36.

Step-by-Step Solution

1
Express the base as a negative remainder modulo the divisor.
When 3636 is divided by 3737, the remainder can be expressed as −1-1 because 36=37×1−136 = 37 \times 1 - 1. Therefore, we can state that 36≡−1(mod37)36 \equiv -1 \pmod{37}.
Using a negative remainder with an absolute value of 11 makes calculating large exponentiations trivial.
2
Apply the exponent to the negative remainder and convert back to a positive value.
Calculate (−1)43(-1)^{43}. Since 4343 is an odd number, (−1)43=−1(-1)^{43} = -1. To find the valid positive remainder, add the divisor: −1+37=36-1 + 37 = 36.
The rules of modular arithmetic allow us to raise the remainder to the given power. However, the final answer must always be a non-negative integer strictly less than the divisor.

Key Concept

Negative Remainders for Large Powers

Alternative Method

You can use Fermat's Little Theorem, which states ap−1≡1(modp)a^{p-1} \equiv 1 \pmod p for any prime pp (as long as aa is not divisible by pp). Here, p=37p = 37, so 3636≡1(mod37)36^{36} \equiv 1 \pmod{37}. We can split the expression: 3643=3636×367≡1×367(mod37)36^{43} = 36^{36} \times 36^7 \equiv 1 \times 36^7 \pmod{37}. Using the negative remainder trick for the remaining part gives (−1)7=−1(-1)^7 = -1. Adding 3737 yields the positive remainder of 3636.
Estimated Time:30s
Question 3Question →

What is the unit digit of the expression 743+9277^{43} + 9^{27}?

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Answer: 2

Answer

The unit digit of the given expression 743+9277^{43} + 9^{27} is 2.
The unit digit of 7437^{43} is determined by dividing 43 by 4, giving a remainder of 3, so 737^3 ends in 3. The unit digit of 9279^{27} is determined by taking 9 to an odd power, which ends in 9. Adding 3+9=123 + 9 = 12 gives a final unit digit of 2.

Step-by-Step Solution

1
Determine the unit digit of 7437^{43} using cyclicity rules.
The unit digit of 7437^{43} is 3.
The base 7 follows a 4-step cyclicity pattern (7, 9, 3, 1). Dividing exponent 43 by 4 leaves remainder 3, giving 73(mod10)=37^3 \pmod{10} = 3.
2
Determine the unit digit of 9279^{27} using cyclicity rules.
The unit digit of 9279^{27} is 9.
The base 9 follows a 2-step cyclicity pattern (9 for odd powers, 1 for even powers). Since 27 is odd, the unit digit is 9.
3
Combine the resulting unit digits.
The unit digit of the sum is 2.
Summing the individual unit digits gives 3+9=123 + 9 = 12, whose unit digit is 2.

Key Concept

Unit Digit and Cyclicity
Question 4Question →
What is the numerical value of the following mathematical expression evaluated using the BODMAS rule?
12.5% of 480−[334÷{1.25+12 of (3.6−1.8−0.7‾)}]×81512.5\% \text{ of } 480 - \left[ 3 \frac{3}{4} \div \left\{ 1.25 + \frac{1}{2} \text{ of } \left( 3.6 - \overline{1.8 - 0.7} \right) \right\} \right] \times \frac{8}{15}
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Answer: 59.2

Answer

The evaluated numerical value of the expression is 59.2.
Following strict BODMAS hierarchy—evaluating the vinculum (1.1), parentheses (2.5), 'of' operation (1.25), curly brackets (2.5), square bracket division (1.5), multiplication (0.8), percentage term (60), and final subtraction—gives 60 - 0.8 = 59.2.

Step-by-Step Solution

1
Evaluate the expression under the vinculum bar.
1.8 - 0.7 = 1.1
The vinculum has the highest priority and acts as an implicit innermost bracket.
2
Evaluate the terms inside the round brackets (parentheses).
3.6 - 1.1 = 2.5
Round brackets must be resolved after the vinculum.
3
Evaluate the 'of' operation inside the curly brackets.
(1/2) * 2.5 = 1.25
'Of' precedes addition and multiplication in the standard BODMAS order.
4
Complete the addition within the curly brackets.
1.25 + 1.25 = 2.5 = 5/2
Brackets must be fully simplified from inside out.
5
Perform division inside the square brackets.
(15/4) / (5/2) = (15/4) * (2/5) = 3/2 = 1.5
Square brackets are evaluated after inner curly brackets.
6
Multiply the square bracket result by 8/15.
1.5 * (8/15) = (3/2) * (8/15) = 4/5 = 0.8
Multiplication takes precedence over final subtraction.
7
Calculate the initial percentage 'of' term.
12.5% of 480 = 0.125 * 480 = 60
'Of' operation on percentage takes precedence before subtraction.
8
Subtract the bracketed product from the initial term.
60 - 0.8 = 59.2
Subtraction is performed last according to BODMAS.

Key Concept

BODMAS Order of Operations with Vinculum and Nested Brackets
Estimated Time:2m 0s
Question 5Question →

What is the unit digit of the expression S=(1!+2!+3!+⋯+99!)+399S = (1! + 2! + 3! + \dots + 99!) + 3^{99}?

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Answer: 0

Answer

The unit digit of the given expression is 0.
The unit digit of a sum is determined by the sum of the unit digits of its individual terms. For the factorial sum (1!+2!+3!+⋯+99!)(1! + 2! + 3! + \dots + 99!), terms from 5!5! onward end in 0 because 5!=1205! = 120. Summing the first four terms yields 1+2+6+24=331 + 2 + 6 + 24 = 33, contributing a unit digit of 3. For 3993^{99}, the cyclicity sequence of base 3 is 3, 9, 7, 1 (length 4). Dividing 99 by 4 leaves a remainder of 3, corresponding to 33=273^3 = 27, which contributes a unit digit of 7. Summing the two unit digits yields 3+7=103 + 7 = 10, giving a final unit digit of 0.

Step-by-Step Solution

1
Evaluate the unit digit contribution of the factorial terms
Unit digit of (1!+2!+3!+⋯+99!)(1! + 2! + 3! + \dots + 99!) is 3
Since n!n! ends in 0 for every n≥5n \ge 5, only 1!+2!+3!+4!=331! + 2! + 3! + 4! = 33 affects the unit digit.
2
Find the unit digit of 3993^{99} using pattern cyclicity
Unit digit of 3993^{99} is 7
The cyclicity of base 3 is 4. Since 99(mod4)=399 \pmod 4 = 3, the unit digit corresponds to 33=273^3 = 27.
3
Combine the unit digits of both parts
Unit digit of SS is 0
Adding the unit digits gives 3+7=103 + 7 = 10, making the final unit digit 0.

Key Concept

Unit digit evaluation using factorial terminal zeros and exponential cyclicity
Question 6Question →
Evaluate the following numerical expression using the correct order of operations (BODMAS):
40−4×(6−2)÷2+540 - 4 \times (6 - 2) \div 2 + 5
What is the final result?
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Answer: 37

Answer

The final result of the expression is 37.
Following the BODMAS rule: first simplify the bracket (6−2)=4(6 - 2) = 4. Next, execute multiplication and division from left to right: 4×4=164 \times 4 = 16 and 16÷2=816 \div 2 = 8. The expression reduces to 40−8+540 - 8 + 5. Finally, carry out subtraction and addition from left to right: 40−8=3240 - 8 = 32, and 32+5=3732 + 5 = 37.

Step-by-Step Solution

1
Evaluate the expression inside the brackets
6−2=46 - 2 = 4
According to BODMAS, operations inside brackets (B) take highest priority.
2
Perform multiplication and division from left to right
4×4=164 \times 4 = 16, then 16÷2=816 \div 2 = 8
Multiplication and Division (DM) carry equal priority and are evaluated in order from left to right.
3
Perform addition and subtraction from left to right
40−8+5=32+5=3740 - 8 + 5 = 32 + 5 = 37
Addition and Subtraction (AS) carry equal priority and are evaluated sequentially from left to right.

Key Concept

BODMAS Order of Operations
Question 7Question →

Let NN be the smallest positive integer that leaves a remainder of 33 when divided by 1111, a remainder of 55 when divided by 1313, and is exactly divisible by 77. What is the remainder when N4+N2+1N^4 + N^2 + 1 is divided by 1717?

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Answer: 3

Answer

The remainder when N4+N2+1N^4 + N^2 + 1 is divided by 1717 is 3.
Using the Chinese Remainder Theorem, the smallest integer meeting all divisibility constraints is determined to be N=707N = 707. Reducing 707707 modulo 1717 gives 1010. Substituting 1010 into the expression N4+N2+1N^4 + N^2 + 1 modulo 1717 yields (104+102+1)(mod17)=(4+15+1)(mod17)=20(mod17)=3(10^4 + 10^2 + 1) \pmod{17} = (4 + 15 + 1) \pmod{17} = 20 \pmod{17} = 3.

Step-by-Step Solution

1
Express the given conditions as a system of linear congruences
N≡3(mod11)N \equiv 3 \pmod{11}, N≡5(mod13)N \equiv 5 \pmod{13}, and N≡0(mod7)N \equiv 0 \pmod{7}.
Translates the problem statement into standard modular arithmetic notation.
2
Solve the simultaneous congruences for modulo 11 and modulo 13
Let N=11k+3N = 11k + 3. Substituting into 11k+3≡5(mod13)11k + 3 \equiv 5 \pmod{13} gives 11k≡2(mod13)  ⟹  −2k≡2(mod13)  ⟹  k≡12(mod13)11k \equiv 2 \pmod{13} \implies -2k \equiv 2 \pmod{13} \implies k \equiv 12 \pmod{13}. Thus, N=11(13m+12)+3=143m+135N = 11(13m + 12) + 3 = 143m + 135.
Combines two modular constraints into a single general expression using the Chinese Remainder Theorem principle.
3
Apply the divisibility condition for 7 to determine the smallest value of NN
143m+135≡0(mod7)  ⟹  3m+2≡0(mod7)  ⟹  3m≡5(mod7)  ⟹  m≡4(mod7)143m + 135 \equiv 0 \pmod{7} \implies 3m + 2 \equiv 0 \pmod{7} \implies 3m \equiv 5 \pmod{7} \implies m \equiv 4 \pmod{7}. For the smallest positive integer, set m=4m = 4, so N=143(4)+135=707N = 143(4) + 135 = 707.
Finds the specific smallest integer NN that satisfies all three divisor requirements.
4
Reduce NN modulo 17
707=17×41+10  ⟹  N≡10(mod17)707 = 17 \times 41 + 10 \implies N \equiv 10 \pmod{17}.
Simplifies the base before evaluating higher powers.
5
Calculate the polynomial expression (N4+N2+1)(mod17)(N^4 + N^2 + 1) \pmod{17}
N2≡102=100≡15≡−2(mod17)N^2 \equiv 10^2 = 100 \equiv 15 \equiv -2 \pmod{17}, and N4≡(−2)2=4(mod17)N^4 \equiv (-2)^2 = 4 \pmod{17}. Therefore, N4+N2+1≡4+15+1=20≡3(mod17)N^4 + N^2 + 1 \equiv 4 + 15 + 1 = 20 \equiv 3 \pmod{17}.
Uses modular exponentiation properties to efficiently evaluate the final remainder.

Key Concept

Chinese Remainder Theorem and Modular Exponentiation
Question 8Question →

Determine the unit digit of the composite exponential expression E=171819+181920−192021E = 17^{18^{19}} + 18^{19^{20}} - 19^{20^{21}}.

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Answer: 8

Answer

8
The unit digit of 17181917^{18^{19}} is 1 because 181918^{19} is divisible by 4, which corresponds to the 4th power 74≡1(mod10)7^4 \equiv 1 \pmod{10}. The unit digit of 18192018^{19^{20}} is 8 because 1920≡1(mod4)19^{20} \equiv 1 \pmod 4, giving 81=88^1 = 8. The unit digit of 19202119^{20^{21}} is 1 because 202120^{21} is an even exponent, giving 9even≡1(mod10)9^{\text{even}} \equiv 1 \pmod{10}. Combining these gives 1+8−1=81 + 8 - 1 = 8.

Step-by-Step Solution

1
Find the unit digit of 17181917^{18^{19}}
The unit digit is 1
Base 7 has cyclicity 4. Exponent 1819=219≡0(mod4)18^{19} = 2^{19} \equiv 0 \pmod 4. When the remainder modulo 4 is 0, we take the 4th power in the cycle: 74≡1(mod10)7^4 \equiv 1 \pmod{10}.
2
Find the unit digit of 18192018^{19^{20}}
The unit digit is 8
Base 8 has cyclicity 4. Exponent 1920≡(−1)20≡1(mod4)19^{20} \equiv (-1)^{20} \equiv 1 \pmod 4. The 1st power in the cycle gives 81=88^1 = 8.
3
Find the unit digit of 19202119^{20^{21}}
The unit digit is 1
Base 9 has cyclicity 2. The exponent 202120^{21} is even, so 9even≡1(mod10)9^{\text{even}} \equiv 1 \pmod{10}.
4
Evaluate the combined expression
The final unit digit is 8
Combine the unit digits of the individual terms: 1+8−1=81 + 8 - 1 = 8.

Key Concept

Cyclicity of numbers and modular arithmetic for tower exponents
Question 9Question →

What is the unit digit of the expression 252+5172^{52} + 5^{17}?

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Answer: 1

Answer

The unit digit of the expression 252+5172^{52} + 5^{17} is 1.
For 2522^{52}, dividing 52 by 4 leaves a remainder of 0. Since the cyclicity of 2 is 4, a remainder of 0 means using 24=162^4 = 16, which has a unit digit of 6. For 5175^{17}, any power of 5 ends in 5. The total unit digit is the unit digit of 6+5=116 + 5 = 11, which is 1.

Step-by-Step Solution

1
Find the unit digit of 2522^{52} using cyclicity.
The cyclicity of 2 is 4 (pattern: 2, 4, 8, 6). Dividing the exponent 52 by 4 yields 52=4×13+052 = 4 \times 13 + 0. A remainder of 0 corresponds to the 4th power in the cycle. 24=162^4 = 16, so the unit digit of 2522^{52} is 6.
When the exponent is exactly divisible by the cyclicity length, the unit digit corresponds to the power equal to the cyclicity length.
2
Find the unit digit of 5175^{17}.
The cyclicity of 5 is 1, meaning any positive integer power of 5 ends in 5. Thus, the unit digit of 5175^{17} is 5.
Base numbers ending in 5 always yield a unit digit of 5 for any positive integer exponent.
3
Add the individual unit digits to get the final unit digit.
Unit digit of (252+517)=Unit digit of (6+5)=Unit digit of 11=1(2^{52} + 5^{17}) = \text{Unit digit of }(6 + 5) = \text{Unit digit of }11 = 1.
The unit digit of a sum is equal to the unit digit of the sum of the individual unit digits.

Key Concept

Unit Digit and Cyclicity
Estimated Time:45s
Question 10Question →
What is the simplified value of the following mathematical expression obtained by applying the BODMAS rule?
96÷[4+{25−(12−8−3‾)×3}]96 \div \left[ 4 + \left\{ 25 - \left( 12 - \overline{8 - 3} \right) \times 3 \right\} \right]
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Answer: 12

Answer

12
Following the standard BODMAS precedence rule: first evaluate the vinculum 8−3‾=5\overline{8-3} = 5, then the round brackets (12−5)=7(12-5) = 7, then multiplication 7×3=217 \times 3 = 21, then curly brackets 25−21=425 - 21 = 4, then square brackets 4+4=84 + 4 = 8, and finally division 96÷8=1296 \div 8 = 12.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
\overline{8 - 3} = 5
The vinculum takes precedence over other bracket operations.
2
Evaluate the operation inside the innermost round brackets
12 - 5 = 7
Resolve the expression within the round brackets next.
3
Perform multiplication inside the curly brackets
7×3=217 \times 3 = 21
Multiplication precedes subtraction within the curly brackets according to BODMAS.
4
Complete the subtraction inside the curly brackets
25 - 21 = 4
Evaluate the remaining subtraction inside the curly brackets.
5
Evaluate the addition inside the outer square brackets
4 + 4 = 8
Combine the terms within the square brackets to obtain the final divisor.
6
Perform the final division
96÷8=1296 \div 8 = 12
Divide the initial integer by the evaluated square bracket result.

Key Concept

Simplification and BODMAS Rule with Vinculum
Question 11Question →

What is the unit digit of the expression E=(842324×617153)+358217E = (842^{324} \times 617^{153}) + 358^{217}?

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Answer: 0

Answer

The unit digit of the expression E=(842324×617153)+358217E = (842^{324} \times 617^{153}) + 358^{217} is 0.
For 842324842^{324}, exponent 324(mod4)=0324 \pmod 4 = 0 corresponds to the 4th power position in cyclicity (24=16→62^4 = 16 \rightarrow 6). For 617153617^{153}, 153(mod4)=1153 \pmod 4 = 1 corresponds to 71=77^1 = 7. The product of these terms ends in 6×7=42→26 \times 7 = 42 \rightarrow 2. For 358217358^{217}, 217(mod4)=1217 \pmod 4 = 1 corresponds to 81=88^1 = 8. Adding the product's unit digit (2) and 358217358^{217}'s unit digit (8) gives 2+8=10→02 + 8 = 10 \rightarrow 0.

Step-by-Step Solution

1
Find the unit digit of 842324842^{324}
Unit digit is 6
The unit digit of base 842 is 2, which repeats unit digits in a cycle of 4: (2,4,8,6)(2, 4, 8, 6). Dividing exponent 324 by 4 yields remainder 0, indicating the 4th position in the cycle, 24=16→62^4 = 16 \rightarrow 6.
2
Find the unit digit of 617153617^{153}
Unit digit is 7
The unit digit of base 617 is 7, which repeats in a cycle of 4: (7,9,3,1)(7, 9, 3, 1). Dividing exponent 153 by 4 yields remainder 1, indicating the 1st position in the cycle, 71=77^1 = 7.
3
Calculate the unit digit of the product (842324×617153)(842^{324} \times 617^{153})
Unit digit is 2
Multiplying the unit digits of the two terms gives 6×7=426 \times 7 = 42, which has a unit digit of 2.
4
Find the unit digit of 358217358^{217}
Unit digit is 8
The unit digit of base 358 is 8, which repeats in a cycle of 4: (8,4,2,6)(8, 4, 2, 6). Dividing exponent 217 by 4 yields remainder 1, indicating the 1st position in the cycle, 81=88^1 = 8.
5
Add the unit digit of the product to the unit digit of 358217358^{217}
Final unit digit is 0
Adding the unit digit of the product (2) and the unit digit of the final term (8) gives 2+8=102 + 8 = 10, which ends in 0.

Key Concept

Unit Digit Computation using Exponents and Cyclicity
Question 12Question →
What is the simplified numerical value of the following mathematical expression when evaluated using the standard BODMAS rule?
240÷[212×{14−(312+25% of (14−1.8+2.2‾))}]240 \div \left[ 2 \frac{1}{2} \times \left\{ 14 - \left( 3 \frac{1}{2} + 25\% \text{ of } \left( 14 - \overline{1.8 + 2.2} \right) \right) \right\} \right]
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Answer: 12

Answer

12
Following the strict hierarchical order of operations (BODMAS: Vinculum →\rightarrow Brackets →\rightarrow Orders →\rightarrow Division/Multiplication →\rightarrow Addition/Subtraction), the innermost vinculum 1.8+2.2‾\overline{1.8 + 2.2} evaluates to 44. Then 14−4=1014 - 4 = 10, and 25%25\% of 1010 equals 2.52.5. Adding this to 312(3.5)3 \frac{1}{2} (3.5) yields 66. Inside the curly brackets, 14−6=814 - 6 = 8. Inside the square brackets, 212(2.5)×8=202 \frac{1}{2} (2.5) \times 8 = 20. Finally, 240÷20=12240 \div 20 = 12.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
\overline{1.8 + 2.2} = 4
The vinculum acts as the innermost grouping symbol and must be simplified first.
2
Simplify the innermost parenthetical difference
14 - 4 = 10
Subtract the vinculum result from 14 within the round brackets.
3
Calculate the percentage of the result from Step 2
25\% \text{ of } 10 = \frac{25}{100} \times 10 = 2.5
'Of' represents multiplication in percentage evaluation.
4
Complete the evaluation inside the round brackets
3 \frac{1}{2} + 2.5 = 3.5 + 2.5 = 6
Convert the mixed fraction 3123 \frac{1}{2} to decimal 3.53.5 and sum with 2.52.5.
5
Evaluate the expression inside the curly brackets
14 - 6 = 8
Subtract the round bracket total from 14.
6
Simplify the expression inside the square brackets
2 \frac{1}{2} \times 8 = 2.5 \times 8 = 20
Multiply the mixed fraction 2122 \frac{1}{2} by the curly bracket result.
7
Perform the final division outside the brackets
240÷20=12240 \div 20 = 12
Divide 240 by the completely simplified value of the square brackets.

Key Concept

BODMAS Rule with Vinculum and Multi-level Brackets
Question 13Question →

What is the remainder when the expression 15100+1910015^{100} + 19^{100} is divided by 1717?

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Answer: 15

Answer

The remainder when 15100+1910015^{100} + 19^{100} is divided by 1717 is 1515.
Reducing the bases modulo 17 gives 15≡−215 \equiv -2 and 19≡219 \equiv 2. Since 100100 is even, (−2)100+2100=2⋅2100=2101(-2)^{100} + 2^{100} = 2 \cdot 2^{100} = 2^{101}. By Fermat's Little Theorem (216≡1(mod17)2^{16} \equiv 1 \pmod{17}), 2101≡25=32≡15(mod17)2^{101} \equiv 2^5 = 32 \equiv 15 \pmod{17}. Thus, the correct remainder is 15.

Step-by-Step Solution

1
Reduce the base numbers modulo 17
15≡−2(mod17)15 \equiv -2 \pmod{17} and 19≡2(mod17)19 \equiv 2 \pmod{17}
Using negative remainders simplifies power calculations.
2
Rewrite the expression using modular congruence
15100+19100≡(−2)100+2100=2100+2100=2×2100=2101(mod17)15^{100} + 19^{100} \equiv (-2)^{100} + 2^{100} = 2^{100} + 2^{100} = 2 \times 2^{100} = 2^{101} \pmod{17}
Since the exponent 100100 is even, (−2)100=2100(-2)^{100} = 2^{100}.
3
Apply Fermat's Little Theorem to simplify the exponent
216≡1(mod17)2^{16} \equiv 1 \pmod{17}
17 is prime and gcd⁡(2,17)=1\gcd(2, 17) = 1.
4
Divide the exponent 101 by 16 to find the remaining power
101=16×6+5  ⟹  2101≡(216)6×25≡16×32≡32(mod17)101 = 16 \times 6 + 5 \implies 2^{101} \equiv (2^{16})^6 \times 2^5 \equiv 1^6 \times 32 \equiv 32 \pmod{17}
Any integer power of 1 is 1 under modular arithmetic.
5
Compute the final positive remainder
32=17×1+15  ⟹  32≡15(mod17)32 = 17 \times 1 + 15 \implies 32 \equiv 15 \pmod{17} (or −2+17=15-2 + 17 = 15)
Remainders must be non-negative integers strictly less than the divisor.

Key Concept

Fermat's Little Theorem and Negative Remainders in Modular Arithmetic
Question 14Question →

What is the final value of the expression 36÷4×3+(15−7)−1036 \div 4 \times 3 + (15 - 7) - 10?

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Answer: 25

Answer

The correct answer is 25.
Applying the BODMAS order of operations strictly: first simplify the bracket (15−7)=8(15 - 7) = 8. Next, evaluate multiplication and division from left to right: 36÷4=936 \div 4 = 9 and 9×3=279 \times 3 = 27. Finally, evaluate addition and subtraction left to right: 27+8−10=2527 + 8 - 10 = 25.

Step-by-Step Solution

1
Evaluate bracket expression
15 - 7 = 8
Brackets have the highest precedence in the BODMAS rule.
2
Perform division and multiplication from left to right
36 ÷ 4 = 9, then 9 × 3 = 27
Division and multiplication carry equal priority and are evaluated in order from left to right.
3
Perform addition and subtraction from left to right
27 + 8 - 10 = 25
Addition and subtraction are performed sequentially from left to right.

Key Concept

BODMAS Order of Operations
Estimated Time:45s
Question 15Question →

In basic arithmetic, the unit digit of a power of an integer repeats in a periodic pattern known as cyclicity. What is the unit digit of 3453^{45}?

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Answer: 3

Answer

The unit digit of 3453^{45} is 3.
The base number 3 has a cyclicity pattern of length 4: (3, 9, 7, 1). When the exponent 45 is divided by 4, the quotient is 11 and the remainder is 1. A remainder of 1 indicates the unit digit is equal to the first element of the cyclic pattern, which is 3.

Step-by-Step Solution

1
Identify the unit digit cyclicity pattern for base 3.
The unit digits repeat every 4 powers in the pattern (3, 9, 7, 1).
Calculating sequential powers of 3 yields last digits of 3, 9, 7, 1, 3, 9, 7, 1, establishing a cycle length of 4.
2
Divide the exponent by the cycle period of 4.
45÷4=1145 \div 4 = 11 with a remainder of 11.
The remainder indicates the exact position within the repeating cycle.
3
Find the unit digit using the remainder.
Since the remainder is 1, the unit digit is 31=33^1 = 3.
The first number in the cyclicity sequence (3, 9, 7, 1) is 3.

Key Concept

Unit Digit Cyclicity
Estimated Time:45s
Question 16Question →
Evaluate the following mathematical expression using the standard BODMAS rule:
40% of 150+[72÷{18−(212×4+11−5‾)}]40\% \text{ of } 150 + \left[ 72 \div \left\{ 18 - \left( 2\frac{1}{2} \times 4 + \overline{11 - 5} \right) \right\} \right]
What is the final simplified numerical value?
Show answer & explanation

Answer: 96

Answer

The simplified numerical value of the given expression is 96.
Applying the standard BODMAS rule: first evaluate the vinculum 11−5‾=6\overline{11 - 5} = 6. Next, evaluate inside the round brackets: 2.5×4+6=162.5 \times 4 + 6 = 16. Then simplify inside the curly braces: 18−16=218 - 16 = 2. Following that, perform division inside the square brackets: 72÷2=3672 \div 2 = 36. Separately, evaluate the percentage term: 40% of 150=6040\% \text{ of } 150 = 60. Summing these two terms gives 60+36=9660 + 36 = 96.

Step-by-Step Solution

1
Evaluate the bar (vinculum) expression.
\overline{11 - 5} = 6
The vinculum (bar) has highest priority within bracket operations.
2
Simplify the terms within the round parentheses.
2\frac{1}{2} \times 4 + 6 = 10 + 6 = 16
Perform multiplication before addition inside parentheses.
3
Evaluate the expression inside the curly braces.
18 - 16 = 2
Subtract the result of the round brackets from 18.
4
Perform the division inside the square brackets.
72÷2=3672 \div 2 = 36
Divide 72 by the value obtained from curly braces.
5
Calculate the percentage component of the expression.
40\% \text{ of } 150 = \frac{40}{100} \times 150 = 60
The word 'of' indicates multiplication applied to the percentage value.
6
Perform the final addition.
60 + 36 = 96
Add the result of the percentage calculation to the result of the bracket evaluation.

Key Concept

Order of Operations (BODMAS Rule) with Vinculum, Nested Brackets, and Percentages
Estimated Time:1m 30s
Question 17Question →
Evaluate the following mathematical expression by strictly applying the BODMAS rule:
150% of 48−[14.5−{412×(12.8÷2.5+1.5‾−135)+2.8}]÷0.15150\% \text{ of } 48 - \left[ 14.5 - \left\{ 4 \frac{1}{2} \times \left( 12.8 \div \overline{2.5 + 1.5} - 1 \frac{3}{5} \right) + 2.8 \right\} \right] \div 0.15
What is the simplified numerical value?
Show answer & explanation

Answer: 42

Answer

The simplified numerical value of the given mathematical expression is 42.
Following the strict BODMAS order of operations: Vinculum -> Round brackets -> Curly brackets -> Square brackets -> Division -> 'Of' -> Subtraction gives the exact result 42.

Step-by-Step Solution

1
Evaluate the term under the vinculum (bar)
2.5 + 1.5 = 4
The vinculum has highest priority among grouping symbols.
2
Evaluate operations inside the round brackets (parentheses)
12.8 / 4 - 1.6 = 3.2 - 1.6 = 1.6
Division inside parentheses precedes subtraction, and 1 3/5 converts to decimal 1.6.
3
Evaluate operations inside the curly brackets
4.5 * 1.6 + 2.8 = 7.2 + 2.8 = 10
Multiplication precedes addition inside curly brackets, with 4 1/2 converting to 4.5.
4
Evaluate operations inside the square brackets
14.5 - 10 = 4.5
Subtract the evaluated inner expression from 14.5.
5
Perform the division following the bracket result
4.5 / 0.15 = 30
Division operation takes precedence over final subtraction.
6
Calculate the percentage 'of' expression
150% of 48 = (150 / 100) * 48 = 72
'Of' operation is evaluated before basic addition and subtraction.
7
Compute the final subtraction
72 - 30 = 42
Subtract the result of the bracketed division term from the percentage evaluation.

Key Concept

Order of Operations (BODMAS / PEMDAS) with Vinculum and Nested Brackets
Question 18Question →
Find the value of the given mathematical expression by applying the standard order of operations (BODMAS):
25−[12+{8÷2×(5−3+1‾)}]25 - [12 + \{8 \div 2 \times (5 - \overline{3 + 1})\}]
Show answer & explanation

Answer: 9

Answer

9
Following the BODMAS rule: evaluate the vinculum 3+1‾=4\overline{3 + 1} = 4, then round brackets (5−4)=1(5 - 4) = 1, then curly brackets {8÷2×1}=4\{8 \div 2 \times 1\} = 4, then square brackets [12+4]=16[12 + 4] = 16, and finally 25−16=925 - 16 = 9.

Step-by-Step Solution

1
Evaluate the expression under the vinculum (bar)
3 + 1 = 4
The vinculum has the highest precedence among brackets (VBODMAS).
2
Evaluate the innermost round bracket
5 - 4 = 1
Round brackets are evaluated after resolving the vinculum.
3
Evaluate division and multiplication inside curly brackets from left to right
8 ÷ 2 × 1 = 4 × 1 = 4
Division and multiplication have equal precedence and are resolved from left to right.
4
Evaluate addition inside square brackets
12 + 4 = 16
Square brackets are resolved next.
5
Perform final subtraction outside the brackets
25 - 16 = 9
Outer subtraction is performed as the final step.

Key Concept

Order of Operations (BODMAS / VBODMAS)
Estimated Time:45s
Question 19Question →

What is the unit digit of the composite exponential expression E=(432316×657235)+875432−959411E = (432^{316} \times 657^{235}) + 875^{432} - 959^{411}?

Show answer & explanation

Answer: 4

Answer

The unit digit of the expression is 4.
Evaluating each term using cyclicity: 432316432^{316} ends in 6 (242^4), 657235657^{235} ends in 3 (737^3), 875432875^{432} ends in 5 (5n5^n), and 959411959^{411} ends in 9 (9odd9^{\text{odd}}). Combining these gives (6×3)+5−9=18+5−9=14(6 \times 3) + 5 - 9 = 18 + 5 - 9 = 14, yielding a unit digit of 4.

Step-by-Step Solution

1
Determine the unit digit of 432316432^{316}
Unit digit is 6
The unit digit of the base is 2, which has a cyclicity pattern of 4 (2, 4, 8, 6). The exponent 316 is divisible by 4 with remainder 0. For a remainder of 0, we take the 4th power in the cycle: 24=162^4 = 16, giving a unit digit of 6.
2
Determine the unit digit of 657235657^{235}
Unit digit is 3
The unit digit of the base is 7, which has a cyclicity pattern of 4 (7, 9, 3, 1). Dividing the exponent 235 by 4 gives a remainder of 3 (235=4×58+3235 = 4 \times 58 + 3). We take the 3rd power in the cycle: 73=3437^3 = 343, giving a unit digit of 3.
3
Calculate the unit digit of the product (432316×657235)(432^{316} \times 657^{235})
Unit digit is 8
Multiplying the unit digits of the two terms yields 6×3=186 \times 3 = 18, which has a unit digit of 8.
4
Determine the unit digit of 875432875^{432} and add it to the product
Unit digit of sum is 3
Any positive integer power of a base ending in 5 always ends in 5. Adding this to the product's unit digit gives 8+5=138 + 5 = 13, so the sum ends in 3.
5
Determine the unit digit of 959411959^{411} and subtract it to find the final unit digit
Final unit digit is 4
The unit digit of the base is 9, which has a cyclicity of 2 (9 for odd powers, 1 for even powers). Since 411 is odd, 959411959^{411} ends in 9. Subtracting this from the sum's unit digit gives 3−9≡13−9=43 - 9 \equiv 13 - 9 = 4.

Key Concept

Unit digit calculation using cyclicity patterns and modular arithmetic for multi-term exponential expressions.
Question 20Question →

A positive integer NN when successively divided by 44, 55, and 66 leaves remainders of 22, 33, and 44, respectively. What is the sum of the remainders obtained when the smallest such number NN is successively divided by 66, 55, and 44, in that order?

Show answer & explanation

Answer: 7

Answer

The sum of the remainders obtained when the smallest such number is successively divided by 6, 5, and 4 is 7.
The smallest positive integer NN satisfying the given successive division conditions is 94. Successively dividing 94 by 6, 5, and 4 yields quotients of 15, 3, and 0 with remainders 4, 0, and 3, respectively. The sum of these remainders is 4+0+3=74 + 0 + 3 = 7.

Step-by-Step Solution

1
Formulate the algebraic equations for successive division and calculate the smallest positive value of NN.
N=94N = 94
By definition of successive division, N=4q1+2N = 4q_1 + 2, q1=5q2+3q_1 = 5q_2 + 3, and q2=6q3+4q_2 = 6q_3 + 4. Setting the non-negative integer quotient q3=0q_3 = 0 yields q2=4q_2 = 4, q1=23q_1 = 23, and N=94N = 94.
2
Perform successive division of 94 by the divisors 6, 5, and 4 in sequence.
The sequence of remainders is 44, 00, and 33.
First stage: 94÷6=1594 \div 6 = 15 remainder 44. Second stage: 15÷5=315 \div 5 = 3 remainder 00. Third stage: 3÷4=03 \div 4 = 0 remainder 33.
3
Calculate the sum of the three remainders.
4+0+3=74 + 0 + 3 = 7
Adding the individual remainders obtained from each stage of the reverse order division.

Key Concept

Successive Division and Remainder Property
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