Remainders and Units Digit Cyclicity Alıştırma Soruları — GMAT

What is the remainder when the expression $7^{45} \cdot 3^{22} + 8^{15}$ is divided by $10$ ?

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Cevap: 5

Cevap

The remainder is 5.

Dividing any number by 10 leaves a remainder equal to the units digit of that number. By analyzing the units digit pattern (cyclicity of period 4) for powers of 7, 3, and 8:

7^{45} \equiv 7^1 = 7 \pmod{10}

,

3^{22} \equiv 3^2 = 9 \pmod{10}

, and

8^{15} \equiv 8^3 = 2 \pmod{10}

. The expression simplifies to

(7 \times 9) + 2 = 63 + 2 = 65

, which has a units digit of 5. Therefore, the remainder when divided by 10 is 5.

Adım Adım Çözüm

1

Relate remainder modulo 10 to units digit cyclicity.

Finding the remainder when an expression is divided by 10 is equivalent to finding its units digit.

Any positive integer

N

can be expressed as

10k + r

, where

r

is the units digit and the remainder when

N

is divided by 10.

2

Determine the units digit of

7^{45}

.

7^{45} \equiv 7 \pmod{10}

.

Powers of 7 repeat their units digits in a cycle of length 4 (7, 9, 3, 1). Dividing the exponent 45 by 4 gives a remainder of 1, so

7^{45}

has the same units digit as

7^1 = 7

.

3

Determine the units digit of

3^{22}

.

3^{22} \equiv 9 \pmod{10}

.

Powers of 3 repeat their units digits in a cycle of length 4 (3, 9, 7, 1). Dividing the exponent 22 by 4 gives a remainder of 2, so

3^{22}

has the same units digit as

3^2 = 9

.

4

Determine the units digit of the product

7^{45} \cdot 3^{22}

.

7^{45} \cdot 3^{22} \equiv 3 \pmod{10}

.

The product of the units digits is

7 \times 9 = 63

, which has a units digit of 3.

5

Determine the units digit of

8^{15}

.

8^{15} \equiv 2 \pmod{10}

.

Powers of 8 repeat their units digits in a cycle of length 4 (8, 4, 2, 6). Dividing the exponent 15 by 4 gives a remainder of 3, so

8^{15}

has the same units digit as

8^3 = 512

, which ends in 2.

6

Combine the results to find the final remainder modulo 10.

5

Adding the units digit of the first term (3) and the second term (2) gives

3 + 2 = 5

.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic

Soru 2Soru →

A sequence of positive integers $a_n$ is defined by $a_n = 2^n + 5^n$ for all integers $n \ge 1$ . What is the remainder when $a_{100}$ is divided by $7$ ?

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Cevap: 4

Cevap

The remainder when

a_{100}

is divided by

7

is

4

.

By reducing

5 \equiv -2 \pmod 7

, we observe that

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even. The powers of

2 \pmod 7

follow a 3-step cycle (

2, 4, 1

). Dividing the exponent

100

by

3

leaves a remainder of

1

, meaning

2^{100} \equiv 2^1 = 2 \pmod 7

. Adding the remainders for both terms yields

2 + 2 = 4

.

Adım Adım Çözüm

1

Use modular arithmetic to simplify the base

5 \pmod 7

.

5 \equiv -2 \pmod 7

, which implies

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even.

Converting

5

to

-2

allows both terms to be expressed using powers of

2

.

2

Determine the remainder cycle of powers of

2

when divided by

7

.

2^1 \equiv 2 \pmod 7

,

2^2 \equiv 4 \pmod 7

, and

2^3 = 8 \equiv 1 \pmod 7

. The pattern repeats every

3

powers.

Finding the period of cyclicity simplifies evaluating large powers.

3

Divide the exponent

100

by the cycle length

3

.

100 = 3 \times 33 + 1

, leaving a remainder of

1

. Thus,

2^{100} \equiv 2^1 = 2 \pmod 7

.

The remainder of the exponent modulo the cycle length determines the equivalent reduced power.

4

Combine the remainders for

2^{100}

and

5^{100}

.

a_{100} = 2^{100} + 5^{100} \equiv 2 + 2 = 4 \pmod 7

.

Adding the individual modular values gives the overall remainder.

Anahtar Kavram

Modular arithmetic cyclicity and negative congruences

Soru 3Soru →

What is the units digit of $7^{43}$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The units digit of

7^{43}

is 3.

The units digits of powers of 7 repeat in a pattern of four terms: 7, 9, 3, 1. Dividing the exponent 43 by 4 yields a remainder of 3. The 3rd term in the repeating pattern is 3, so the units digit of

7^{43}

is 3.

Adım Adım Çözüm

1

Determine the cyclicity pattern of the units digit of powers of 7.

The units digits for

7^1, 7^2, 7^3, 7^4, \dots

follow the repeating sequence 7, 9, 3, 1 with a cycle length of 4.

Units digits of positive integer powers follow a periodic pattern.

2

Divide the exponent 43 by the pattern cycle length of 4.

43 \div 4 = 10

remainder 3.

The remainder determines which term in the repeating sequence gives the units digit.

3

Find the units digit corresponding to the 3rd term in the cyclicity sequence.

The 3rd digit in the sequence (7, 9, 3, 1) is 3.

A remainder of 3 corresponds to the 3rd power in the cycle,

7^3

.

Anahtar Kavram

Units Digit Cyclicity

Soru 4Soru →

What is the units digit of $8^{21}$ ?

8

4

2

6

0

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

The units digits of integer powers of 8 follow a repeating pattern of length 4: 8, 4, 2, 6. Dividing the exponent 21 by 4 yields

21 = 4 \times 5 + 1

, giving a remainder of 1. A remainder of 1 means the units digit is the first element of the cycle, which is 8.

Adım Adım Çözüm

1

Determine the units digit pattern for powers of 8.

The sequence of units digits for

8^1, 8^2, 8^3, 8^4, \dots

is

8, 4, 2, 6, 8, 4, 2, 6, \dots

, which repeats every 4 powers.

Units digits of positive integer powers follow a periodic cyclic pattern.

2

Find the remainder when the exponent 21 is divided by the cycle length 4.

21 \div 4 = 5

with a remainder of 1.

The remainder determines the position of the units digit within the 4-step cycle.

3

Identify the units digit corresponding to a remainder of 1.

The first number in the pattern

8, 4, 2, 6

is 8.

A remainder of 1 points to the 1st term of the repeating sequence.

Anahtar Kavram

Units Digit Cyclicity

Soru 5Soru →

What is the remainder when the sum $S = 3^1 + 3^2 + 3^3 + \dots + 3^{2025}$ is divided by $100$ ?

Cevabı ve açıklamayı göster

Cevap: 63

Cevap

The remainder when the sum

S

is divided by

100

is

63

.

Using the geometric series formula,

S = \frac{3(3^{2025} - 1)}{2}

. Computing

3^{2025} \pmod{200}

gives

43 \pmod{200}

because

3^{20} \equiv 1 \pmod{200}

and

2025 \equiv 5 \pmod{20}

. Substituting this into the numerator yields

3(43 - 1) = 126 \pmod{200}

. Dividing by

2

produces the remainder

63

modulo

100

.

Adım Adım Çözüm

1

Sum the geometric series

S = 3^1 + 3^2 + \dots + 3^{2025}

S = \frac{3(3^{2025} - 1)}{2}

Applying the standard finite geometric series formula

a\frac{r^n - 1}{r - 1}

with

a = 3

and

r = 3

.

2

Set up modular equation for

S \pmod{100}

Evaluate

3(3^{2025} - 1) \pmod{200}

Dividing an integer congruence by

2

requires doubling the modulus to

200

to maintain exact divisibility.

3

Find the cyclicity period of

3^n \pmod{200}

3^{20} \equiv 1 \pmod{200}

3^5 = 243 \equiv 43

,

3^{10} \equiv 43^2 = 1849 \equiv 49

, and

3^{20} \equiv 49^2 = 2401 \equiv 1 \pmod{200}

.

4

Compute

3^{2025} \pmod{200}

3^{2025} \equiv 3^5 \equiv 43 \pmod{200}

Dividing the exponent

2025

by the period length

20

leaves a remainder of

5

.

5

Calculate the numerator

3(3^{2025} - 1) \pmod{200}

and divide by 2

3(43 - 1) = 126 \pmod{200}

, so

S \equiv 63 \pmod{100}

Simplifying

126 / 2 = 63

gives the unique remainder modulo

100

.

Anahtar Kavram

Geometric series summation combined with modular arithmetic and cyclicity mod 200

Soru 6Soru →

Let $n = 13^{25} - 7^{25}$ . When $n$ is divided by $10$ , the remainder is $R_1$ , where $0 \le R_1 < 10$ . When $-n$ is divided by $7$ , the remainder is $R_2$ , where $0 \le R_2 < 7$ . What is the value of $R_1 + R_2$ ?

5

7

10

12

2

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

R_1 + R_2

is 7.

To find

R_1

, we evaluate

n = 13^{25} - 7^{25} \pmod{10}

. Both 3 and 7 have units digit cyclicity of period 4. Because

25 \equiv 1 \pmod 4

,

13^{25} \equiv 3^1 = 3 \pmod{10}

and

7^{25} \equiv 7^1 = 7 \pmod{10}

. Then

n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}

, so

R_1 = 6

. To find

R_2

, we evaluate

-n = 7^{25} - 13^{25} \pmod 7

. Since

7^{25} \equiv 0 \pmod 7

and

13 \equiv -1 \pmod 7

, we have

-n \equiv 0 - (-1)^{25} = 1 \pmod 7

, so

R_2 = 1

. Summing them yields

R_1 + R_2 = 6 + 1 = 7

.

Adım Adım Çözüm

1

Determine

R_1

, the remainder when

n = 13^{25} - 7^{25}

is divided by 10.

R_1 = 6

Working modulo 10,

13 \equiv 3 \pmod{10}

, so

13^{25} \equiv 3^{25} \pmod{10}

. Powers of 3 follow a units digit cycle of period 4 (

3, 9, 7, 1

). Since

25 = 4 \times 6 + 1

,

3^{25} \equiv 3^1 = 3 \pmod{10}

. Similarly, powers of 7 follow a units digit cycle of period 4 (

7, 9, 3, 1

), so

7^{25} \equiv 7^1 = 7 \pmod{10}

. Thus,

n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}

, yielding

R_1 = 6

.

2

Determine

R_2

, the remainder when

-n = 7^{25} - 13^{25}

is divided by 7.

R_2 = 1

Working modulo 7,

7^{25} \equiv 0 \pmod 7

. Since

13 \equiv -1 \pmod 7

,

13^{25} \equiv (-1)^{25} = -1 \pmod 7

. Substituting these gives

-n = 7^{25} - 13^{25} \equiv 0 - (-1) = 1 \pmod 7

. Since

0 \le 1 < 7

,

R_2 = 1

.

3

Calculate the sum

R_1 + R_2

.

6 + 1 = 7

Adding the two non-negative remainders gives

R_1 + R_2 = 6 + 1 = 7

.

Anahtar Kavram

Modular arithmetic, units digit cyclicity, and converting negative integer remainders to standard non-negative remainders.

Soru 7Soru →

For any positive integer $n$ , let $S_n$ denote the units digit of the sum $2^n + 3^n + 4^n + 7^n$ . What is the remainder when the sum $T = \sum_{n=1}^{102} S_n$ is divided by $9$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The remainder when the sum T is divided by 9 is 1.

Evaluating the units digits of each exponent term reveals that

S_n

follows a repeating 4-term cycle

(6, 8, 2, 4)

with a sum of 20 per cycle. For 102 terms, there are 25 full cycles plus the first two terms (

S_1=6

and

S_2=8

). The total sum is

T = 25 \times 20 + 6 + 8 = 514

. Dividing 514 by 9 yields

514 = 9 \times 57 + 1

, so the remainder is 1.

Adım Adım Çözüm

1

Find the cyclicity of the units digit of each term

2^n, 3^n, 4^n,

and

7^n

.

Units digits repeat in patterns of length 4: for

2^n

(2, 4, 8, 6), for

3^n

(3, 9, 7, 1), for

4^n

(4, 6, 4, 6), and for

7^n

(7, 9, 3, 1).

Units digits of positive integer powers cycle with periodicities that divide 4.

2

Compute

S_n

for

n = 1, 2, 3, 4

and find the sum of one 4-term period.

S_1 = \text{units}(2+3+4+7=16) = 6

,

S_2 = \text{units}(4+9+6+9=28) = 8

,

S_3 = \text{units}(8+7+4+3=22) = 2

,

S_4 = \text{units}(6+1+6+1=14) = 4

. Sum of one period =

6+8+2+4 = 20

.

The sum of the units digits of individual terms determines the units digit of the total expression.

3

Calculate the total sum

T = \sum_{n=1}^{102} S_n

.

Since

102 = 25 \times 4 + 2

, the sequence consists of 25 complete cycles of 4 terms plus the first 2 terms (

S_1=6

and

S_2=8

). Thus,

T = 25 \times 20 + 6 + 8 = 500 + 14 = 514

.

Dividing the total number of terms by the period length gives the number of full cycles and remaining initial terms.

4

Compute the remainder when

T = 514

is divided by 9.

The sum of the digits of 514 is

5 + 1 + 4 = 10

, and

10 \equiv 1 \pmod 9

. Alternatively,

514 = 9 \times 57 + 1

. Thus, the remainder is 1.

A positive integer and the sum of its digits leave the same remainder when divided by 9.

Anahtar Kavram

Units digit cyclicity of exponential terms and modular arithmetic on sequence sums

Tahmini Süre:2m 0s

Soru 8Soru →

If $n$ is a positive integer such that the units digit of $7^n$ is $3$ , what is the remainder when $3^{n+2} + 8^{n+1}$ is divided by $5$ ?

0

1

2

3

4

Cevabı ve açıklamayı göster

Cevap: $4$

Cevap

The correct remainder is

4

.

The cyclicity pattern of the units digit of

7^n

repeats every

4

terms:

7, 9, 3, 1

. For the units digit to be

3

,

n

must leave a remainder of

3

when divided by

4

(

n \equiv 3 \pmod 4

). Using this form,

n+2 \equiv 5 \equiv 1 \pmod 4

, which means

3^{n+2} \equiv 3^1 \equiv 3 \pmod 5

. Similarly,

n+1 \equiv 4 \equiv 0 \pmod 4

, and since

8 \equiv 3 \pmod 5

, we have

8^{n+1} \equiv 3^0 \equiv 1 \pmod 5

. Adding these values gives

3 + 1 = 4

, so the remainder when divided by

5

is

4

.

Adım Adım Çözüm

1

Determine the remainder of

n

when divided by

4

using the units digit cyclicity of

7^n

.

n \equiv 3 \pmod 4

.

The units digits of powers of

7

follow a repeating pattern of period

4

:

7^1 \rightarrow 7

,

7^2 \rightarrow 9

,

7^3 \rightarrow 3

,

7^4 \rightarrow 1

. Since the units digit of

7^n

is

3

,

n

must be of the form

4k + 3

.

2

Evaluate

3^{n+2} \pmod 5

.

3^{n+2} \equiv 3 \pmod 5

.

Substitute

n = 4k + 3

into the exponent:

n + 2 = 4k + 5 = 4(k+1) + 1

. The powers of

3 \pmod 5

repeat every

4

powers (

3^1 \equiv 3, 3^2 \equiv 4, 3^3 \equiv 2, 3^4 \equiv 1

). Thus,

3^{4k+5} \equiv 3^1 \equiv 3 \pmod 5

.

3

Evaluate

8^{n+1} \pmod 5

.

8^{n+1} \equiv 1 \pmod 5

.

First simplify the base:

8 \equiv 3 \pmod 5

, so

8^{n+1} \equiv 3^{n+1} \pmod 5

. Substitute

n = 4k + 3

:

n + 1 = 4k + 4 = 4(k+1)

. Since the exponent is a multiple of

4

,

3^{4(k+1)} \equiv 3^4 \equiv 1 \pmod 5

.

4

Sum the modular results to find the final remainder modulo

5

.

(3 + 1) \pmod 5 = 4

.

By properties of modular addition,

(3^{n+2} + 8^{n+1}) \pmod 5 \equiv (3 + 1) \pmod 5 = 4

.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic Exponent Rules

Tahmini Süre:2m 0s

Soru 9Soru →

Let $S = 2^{2026} - 3^{2026} + 4^{2026} - 7^{2026}$ . What is the remainder when $S$ is divided by $10$ ?

0

2

4

6

8

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The remainder when

S

is divided by

10

is

2

.

Finding the remainder when an expression is divided by 10 is equivalent to finding the units digit of that expression. The units digits of powers of 2, 3, 4, and 7 repeat in periodic cycles of length 4, 4, 2, and 4, respectively. Since

2026 \equiv 2 \pmod 4

, the units digits correspond to the 2nd term of each cycle:

2^2 \rightarrow 4

,

3^2 \rightarrow 9

,

4^2 \rightarrow 6

, and

7^2 \rightarrow 9

. Evaluating the expression yields

4 - 9 + 6 - 9 = -8

. In modular arithmetic, a negative remainder

-8 \pmod{10}

is equivalent to

-8 + 10 = 2

. Therefore, the value representing 2 is correct.

Adım Adım Çözüm

1

Determine the remainder of each term divided by 10 by finding the units digit cyclicity.

Powers of 2 cycle with period 4 (2, 4, 8, 6). Since

2026 = 4 \times 506 + 2

,

2^{2026} \equiv 2^2 \equiv 4 \pmod{10}

.

Dividing an integer by 10 yields a remainder equal to its units digit.

2

Evaluate the units digits for the remaining terms

3^{2026}

,

4^{2026}

, and

7^{2026}

.

Powers of 3 cycle with period 4 (3, 9, 7, 1);

3^{2026} \equiv 3^2 \equiv 9 \pmod{10}

. Powers of 4 cycle with period 2 (4, 6);

4^{2026} \equiv 4^2 \equiv 6 \pmod{10}

. Powers of 7 cycle with period 4 (7, 9, 3, 1);

7^{2026} \equiv 7^2 \equiv 9 \pmod{10}

.

Each base follows a repeating pattern of units digits when raised to successive positive integer powers.

3

Substitute the congruent remainder values back into the expression for

S

.

S \equiv 4 - 9 + 6 - 9 = -8 \pmod{10}

.

Modular arithmetic operations preserve addition and subtraction equivalences.

4

Convert the negative result to a standard non-negative remainder.

-8 + 10 = 2

. Thus, the remainder is

2

.

By definition, the remainder

r

when an integer is divided by

d

must satisfy

0 \leq r < d

.

Anahtar Kavram

Units Digit Cyclicity and Negative Remainder Rules

Tahmini Süre:2m 30s

Soru 10Soru →

An integer sequence is defined by $T_n = 7^n - (-2)^n$ for all positive integers $n$ . What is the remainder when $T_{40}$ is divided by $5$ ?

1

2

0

4

3

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

The remainder when

T_{40}

is divided by

5

is

0

.

Modulo

5

,

7 \equiv 2

, so

7^{40} \equiv 2^{40}

. Since

40

is an even exponent,

(-2)^{40} = 2^{40}

. Substituting these into the formula yields

T_{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5

. Thus, the remainder is

0

.

Adım Adım Çözüm

1

Reduce the base

7

modulo

5

7 \equiv 2 \pmod 5

, so

7^{40} \equiv 2^{40} \pmod 5

.

Simplifying the base makes modular exponentiation straightforward.

2

Evaluate the negative base term

(-2)^{40}

Since

40

is an even integer,

(-2)^{40} = 2^{40}

.

An even power of a negative number yields a positive result.

3

Compute

T_{40}

modulo

5

T_{40} = 7^{40} - (-2)^{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5

.

Subtracting identical values yields

0

.

Anahtar Kavram

Modular Arithmetic and Exponent Parity

Tahmini Süre:1m 30s

Soru 11Soru →

When the positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the units digit of $9^n + 4^{n+1} - 7^{n+2}$ ?

2

4

6

8

0

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The units digit of the expression is 8.

The correct answer is 8 because evaluating each component using unit digit cyclicity gives

9^n ≡ 9 \pmod{10}

(since

n

is odd),

4^{n+1} ≡ 6 \pmod{10}

(since

n+1

is even), and

7^{n+2} ≡ 7 \pmod{10}

(since

n+2 ≡ 1 \pmod 4

). Combining these yields

(9 + 6 - 7) = 8

.

Adım Adım Çözüm

1

Express

n

using division algorithm and determine its properties.

Since

n = 12k + 7

for some non-negative integer

k

,

n

is odd,

n+1

is even, and

n+2 = 12k + 9

.

Establishing the form of

n

determines the exponents for cyclicity calculations.

2

Find the units digit of

9^n

.

Units digit of

9^n

is 9.

Powers of 9 alternate units digits:

9^1 = 9, 9^2 = 1, 9^3 = 9...

Any odd power of 9 ends in 9. Since

n = 12k+7

is odd,

9^n

ends in 9.

3

Find the units digit of

4^{n+1}

.

Units digit of

4^{n+1}

is 6.

Powers of 4 alternate units digits:

4^1 = 4, 4^2 = 6, 4^3 = 4...

Any even power of 4 ends in 6. Since

n

is odd,

n+1

is even, so

4^{n+1}

ends in 6.

4

Find the units digit of

7^{n+2}

.

Units digit of

7^{n+2}

is 7.

Powers of 7 follow a 4-step cyclicity pattern: 7, 9, 3, 1. The exponent

n+2 = 12k + 9 = 4(3k+2) + 1 ≡ 1 \pmod 4

. Thus,

7^{n+2}

has the same units digit as

7^1

, which is 7.

5

Combine the units digits.

Units digit =

9 + 6 - 7 = 8

.

Adding and subtracting the respective units digits gives

15 - 7 = 8

.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic

Tahmini Süre:2m 0s

Soru 12Soru →

When a positive integer $n$ is divided by $12$ , the remainder is $7$ . What is the remainder when $n^2 + 5n + 11$ is divided by $12$ ?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

11

Since

n

leaves a remainder of

7

when divided by

12

, we can substitute

n \equiv 7 \pmod{12}

directly into

n^2 + 5n + 11

. Evaluating

7^2 + 5(7) + 11

gives

49 + 35 + 11 = 95

. Dividing

95

by

12

gives

7

with a remainder of

11

.

Adım Adım Çözüm

1

Express

n

using modular arithmetic

n \equiv 7 \pmod{12}

A positive integer

n

that leaves a remainder of

7

when divided by

12

can be expressed as

n = 12k + 7

for some non-negative integer

k

.

2

Substitute

n \equiv 7 \pmod{12}

into the target expression

n^2 + 5n + 11 \equiv 7^2 + 5(7) + 11 \pmod{12}

By the algebraic properties of remainders (modular arithmetic), substituting the remainder

7

for

n

yields an equivalent remainder modulo

12

.

3

Evaluate the arithmetic sum

49 + 35 + 11 = 95

Computing

7^2 = 49

,

5 \times 7 = 35

, and adding

11

gives

95

.

4

Find the remainder of

95

modulo

12

95 = 12 \times 7 + 11 \implies 95 \equiv 11 \pmod{12}

Dividing

95

by

12

yields a quotient of

7

and a remainder of

11

.

Anahtar Kavram

Modular Arithmetic and Polynomial Remainders

Soru 13Soru →

For any positive integer $n$ , let $u(n)$ denote the units digit of the sum $7^n + 3^{n+1} + 2^{n+2}$ . What is the remainder when the sum $u(1) + u(2) + u(3) + \dots + u(100)$ is divided by 7?

6

5

2

0

1

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The remainder when the sum is divided by 7 is 6.

Evaluating the units digit

u(n)

for

n = 1, 2, 3, 4

gives

u(1) = 4

,

u(2) = 2

,

u(3) = 6

, and

u(4) = 0

. This pattern of 4 values repeats indefinitely. The sum of one full cycle of 4 terms is

4 + 2 + 6 + 0 = 12

. Since there are 100 terms in total, there are

100 / 4 = 25

full cycles. The total sum is

25 \times 12 = 300

. Dividing 300 by 7 gives a quotient of 42 with a remainder of 6. Thus, the correct choice is 6.

Adım Adım Çözüm

1

Determine the units digit cyclicity for each power term.

The units digits of

7^n

follow the 4-term cycle [7, 9, 3, 1]. The units digits of

3^{n+1}

follow [9, 7, 1, 3]. The units digits of

2^{n+2}

follow [8, 4, 2, 6].

Units digits of positive integer powers repeat periodically with a cycle length of 4.

2

Calculate

u(n)

for the first 4 terms to identify the repeating pattern of

u(n)

.

For

n=1

:

7+9+8 = 24 \implies u(1)=4

. For

n=2

:

9+7+4 = 20 \implies u(2)=0

. For

n=3

:

3+1+2 = 6 \implies u(3)=6

. For

n=4

:

1+3+6 = 10 \implies u(4)=0

. The sequence of

u(n)

is

[4, 0, 6, 0]

.

Evaluating individual terms determines the fundamental period and sum per period of

u(n)

.

3

Sum the values of

u(n)

over one complete cycle of 4 terms.

Sum of one cycle =

4 + 0 + 6 + 0 = 10

.

Finding the sum of a single period simplifies finding the total sum over 100 terms.

4

Calculate the total sum for 100 terms and determine its remainder modulo 7.

Since 100 terms contain

100 / 4 = 25

complete cycles, the total sum

S = 25 \times 10 = 250

. Dividing 250 by 7 gives

250 = 7 \times 35 + 5

? Wait:

u(2) = (9+7+4)

units digit is 0. Let's verify:

7^2=49

(9),

3^3=27

(7),

2^4=16

(6).

9+7+6=22 \implies u(2)=2

. Let's re-verify:

u(1)=4, u(2)=2, u(3)=6, u(4)=0

. Sum per cycle =

4+2+6+0 = 12

. Total sum

S = 25 \times 12 = 300

.

300 = 7 \times 42 + 6

. Remainder is 6.

Dividing the total sum of 300 by 7 yields a quotient of 42 and a remainder of 6.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic Sums

Tahmini Süre:1m 30s

Soru 14Soru →

When a positive integer $n$ is divided by $4$ , the remainder is $3$ . What is the units digit of the expression $M = 3^{n+1} + 8^{2n+1} - 7^{n+2}$ ?

2

4

6

8

0

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The units digit of the expression is 6.

Since

n \equiv 3 \pmod{4}

, we substitute into each exponent:

n+1 \equiv 0 \pmod{4}

,

2n+1 \equiv 3 \pmod{4}

, and

n+2 \equiv 1 \pmod{4}

. The units digit cyclicity patterns give

3^4 \to 1

,

8^3 \to 2

, and

7^1 \to 7

. Evaluating the combined units digits yields

1 + 2 - 7 = -4

. Adding 10 to obtain a valid units digit gives

6

.

Adım Adım Çözüm

1

Express the integer

n

in terms of modular arithmetic

n \equiv 3 \pmod{4}

, which means

n = 4k + 3

for some non-negative integer

k

.

Units digits of powers repeat in cycles of 4, so exponent behavior depends on exponents modulo 4.

2

Determine the units digit of

3^{n+1}

Exponent

n+1 = (4k+3)+1 = 4k+4 \equiv 0 \pmod{4}

. The units digit pattern for 3 is

3, 9, 7, 1

. Since exponent mod 4 is 0 (or 4), the units digit is

1

.

The cyclicity pattern for powers of 3 has period length 4.

3

Determine the units digit of

8^{2n+1}

Exponent

2n+1 = 2(4k+3)+1 = 8k+7 \equiv 3 \pmod{4}

. The units digit pattern for 8 is

8, 4, 2, 6

. For exponent mod 4 equal to 3, the units digit is

2

.

The cyclicity pattern for powers of 8 has period length 4.

4

Determine the units digit of

7^{n+2}

Exponent

n+2 = (4k+3)+2 = 4k+5 \equiv 1 \pmod{4}

. The units digit pattern for 7 is

7, 9, 3, 1

. For exponent mod 4 equal to 1, the units digit is

7

.

The cyclicity pattern for powers of 7 has period length 4.

5

Combine the units digits and resolve negative intermediate values

1 + 2 - 7 = -4 \equiv 6 \pmod{10}

. The units digit is

6

.

Units digits must be non-negative integers from 0 to 9, so a negative result requires adding 10.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic

Tahmini Süre:2m 0s

Soru 15Soru →

When the integer $N = 4^{43} + 9^{43}$ is divided by $7$ , what is the remainder?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The remainder when

N = 4^{43} + 9^{43}

is divided by

7

is

6

.

Reducing

9 \pmod 7

yields

2

, allowing

N

to be expressed as

2^{86} + 2^{43} \pmod 7

. The powers of

2 \pmod 7

repeat every 3 powers

(2, 4, 1)

. Reducing the exponents

86

and

43

modulo 3 gives remainders of

2

and

1

, corresponding to values of

4

and

2

. Their sum

4 + 2 = 6

is the final remainder.

Adım Adım Çözüm

1

Simplify the base modulo 7 and express in powers of 2

9 \equiv 2 \pmod 7

, so

N = 4^{43} + 9^{43} \equiv 2^{86} + 2^{43} \pmod 7

.

Reducing bases modulo 7 simplifies calculating large exponents.

2

Determine the remainder cyclicity of powers of 2 modulo 7

The cycle length is 3 with pattern

(2, 4, 1)

, because

2^1 \equiv 2

,

2^2 \equiv 4

, and

2^3 \equiv 1 \pmod 7

.

Powers of integers modulo a divisor repeat periodically.

3

Evaluate each term using the exponent modulo the cycle length

86 \equiv 2 \pmod 3 \implies 2^{86} \equiv 2^2 = 4 \pmod 7

, and

43 \equiv 1 \pmod 3 \implies 2^{43} \equiv 2^1 = 2 \pmod 7

.

The position in the cyclicity sequence is dictated by the exponent modulo the period length.

4

Add the individual remainders

4 + 2 = 6 \pmod 7

.

The remainder of a sum equals the sum of the individual remainders.

Anahtar Kavram

Modular arithmetic cyclicity of powers and addition of remainders

Soru 16Soru →

What is the remainder when $17^{17^{17}}$ is divided by $7$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The remainder when

17^{17^{17}}

is divided by

7

is

5

.

First, reduce the base modulo

7

:

17 \equiv 3 \pmod 7

, turning the expression into

3^{17^{17}} \pmod 7

. Next, evaluate the pattern of powers of

3 \pmod 7

:

3^1 \equiv 3

,

3^2 \equiv 2

,

3^3 \equiv 6

,

3^4 \equiv 4

,

3^5 \equiv 5

, and

3^6 \equiv 1

, showing a period of

6

. To find which term of the cycle corresponds to the exponent

17^{17}

, evaluate

17^{17} \pmod 6

. Since

17 \equiv -1 \pmod 6

,

17^{17} \equiv (-1)^{17} = -1 \equiv 5 \pmod 6

. Finally, the

5

th term of the cycle gives

3^5 = 243 \equiv 5 \pmod 7

. Therefore, the remainder is

5

.

Adım Adım Çözüm

1

Reduce the base modulo 7

17 \equiv 3 \pmod 7

, so

17^{17^{17}} \equiv 3^{17^{17}} \pmod 7

.

Modular arithmetic permits replacing the base of an exponential expression with its remainder upon division by the modulus.

2

Determine the cyclicity period of powers of 3 modulo 7

The remainders of

3^n \pmod 7

repeat in a cycle of length 6:

(3, 2, 6, 4, 5, 1)

.

By Fermat's Little Theorem,

3^6 \equiv 1 \pmod 7

, meaning the sequence of remainders repeats every 6 integer powers.

3

Evaluate the exponent

17^{17}

modulo the period length 6

17 \equiv -1 \pmod 6

, so

17^{17} \equiv (-1)^{17} = -1 \equiv 5 \pmod 6

.

The position within the 6-term cyclicity pattern depends on the exponent modulo 6. Using negative remainders simplifies calculating odd powers of

-1

.

4

Calculate the final remainder using the 5th position in the cyclicity pattern

3^5 = 243 = 7 \times 34 + 5 \equiv 5 \pmod 7

.

Since the exponent leaves a remainder of 5 when divided by 6, the overall remainder corresponds to

3^5 \pmod 7

.

Anahtar Kavram

Modular cyclicity and tower exponent remainder reduction

Soru 17Soru →

What is the units digit of $2^{35}$ ?

2

4

6

8

0

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The units digit of

2^{35}

is 8.

The units digit of powers of 2 repeats every 4 powers in the sequence 2, 4, 8, 6. Dividing the exponent 35 by 4 yields a quotient of 8 and a remainder of 3. A remainder of 3 corresponds to the third number in the repeating sequence, which is 8.

Adım Adım Çözüm

1

Find the cyclicity pattern of the units digits for powers of 2.

The units digits of

2^1, 2^2, 2^3, 2^4, 2^5, \dots

are

2, 4, 8, 6, 2, \dots

, repeating in a cycle of length 4.

Units digits of powers follow a repeating periodic pattern determined by base arithmetic modulo 10.

2

Divide the exponent by the length of the cycle.

35 \div 4 = 8

with a remainder of

3

.

The remainder indicates how far into the 4-term repeating cycle the exponent 35 reaches.

3

Match the remainder to the corresponding term in the units digit cycle.

A remainder of 3 corresponds to the 3rd term in the cycle

(2, 4, 8, 6)

, which is 8.

The 3rd power in the repeating pattern gives a units digit of

2^3 = 8

.

Anahtar Kavram

Units Digit Cyclicity

Tahmini Süre:1m 0s

Soru 18Soru →

What is the remainder when $3^{42}$ is divided by $5$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The remainder when

3^{42}

is divided by 5 is 4.

The remainders of powers of 3 divided by 5 follow a repeating pattern of length 4: (3, 4, 2, 1). To find the remainder of

3^{42} \div 5

, divide the exponent 42 by 4. Since

42 = 4 \times 10 + 2

, the remainder of the exponent is 2. The 2nd term in the repeating sequence is 4, so

3^{42} \pmod 5 = 4

.

Adım Adım Çözüm

1

Determine the remainder pattern for consecutive powers of 3 divided by 5.

The remainders for

3^1, 3^2, 3^3, 3^4, 3^5

are

3, 4, 2, 1, 3

, establishing a cycle of length 4.

Powers of integers divided by a positive integer yield repeating sequence patterns (cyclicity).

2

Divide the target exponent by the cycle length.

Dividing 42 by 4 gives a quotient of 10 and a remainder of 2.

The remainder indicates the specific term within the 4-step cycle.

3

Evaluate the value corresponding to the 2nd position in the cycle.

The 2nd term in the cycle (3, 4, 2, 1) is 4.

A remainder of 2 in the exponent position corresponds to the same remainder as

3^2

divided by 5.

Anahtar Kavram

Remainders and Units Digit Cyclicity

Soru 19Soru →

In number theory, the units digit of a positive integer raised to successive positive integer powers follows a repeating cyclic pattern. What is the units digit of $4^{25}$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The units digit of

4^{25}

is 4.

The units digit of powers of 4 alternates between 4 (for odd powers) and 6 (for even powers). Because 25 is an odd number,

4^{25}

has a units digit of 4.

Adım Adım Çözüm

1

Identify the units digit pattern for powers of 4

The units digits cycle between 4 (for odd exponents) and 6 (for even exponents), giving a cycle length of 2.

Units digits of positive integer powers follow a periodic pattern determined by the base digit.

2

Determine the parity of the exponent 25

25 is an odd integer (remainder 1 when divided by 2).

The exponent's remainder modulo 2 determines which position in the 2-element cycle [4, 6] the number falls into.

3

Select the corresponding units digit from the cycle

Since 25 is odd, the units digit is 4.

Odd powers of 4 always have a units digit of 4.

Anahtar Kavram

Units Digit Cyclicity

Soru 20Soru →

Let $K = 7^{83} - 4^{42}$ . What is the remainder when $-K$ is divided by $5$ ?

1

2

3

4

0

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

To find the remainder of

-K

divided by 5, first calculate

K \pmod 5

. By analyzing powers of 7 modulo 5,

7^1 \equiv 2, 7^2 \equiv 4, 7^3 \equiv 3, 7^4 \equiv 1

, so

7^{83} = 7^{4(20)+3} \equiv 3 \pmod 5

. For powers of 4 modulo 5,

4^1 \equiv 4, 4^2 \equiv 1

, so

4^{42} \equiv 1 \pmod 5

. Therefore,

K \equiv 3 - 1 = 2 \pmod 5

. Consequently,

-K \equiv -2 \pmod 5

. Adding the divisor 5 gives a non-negative remainder of 3.

Adım Adım Çözüm

1

Find the remainder of

7^{83}

when divided by

5

.

7^1 \equiv 2

,

7^2 \equiv 4

,

7^3 \equiv 3

,

7^4 \equiv 1 \pmod 5

. Since

83 = 4 \times 20 + 3

,

7^{83} \equiv 7^3 \equiv 3 \pmod 5

.

Powers of 7 follow a 4-term repeating remainder cycle modulo 5: (2, 4, 3, 1).

2

Find the remainder of

4^{42}

when divided by

5

.

4^1 \equiv 4

,

4^2 \equiv 1 \pmod 5

. Since 42 is even,

4^{42} \equiv 1 \pmod 5

.

Powers of 4 follow a 2-term repeating remainder cycle modulo 5: (4, 1).

3

Compute

K \pmod 5

.

K \equiv 3 - 1 = 2 \pmod 5

.

Modular subtraction allows evaluating terms independently.

4

Determine the positive remainder of

-K \pmod 5

.

-K \equiv -2 \equiv -2 + 5 = 3 \pmod 5

.

Standard division algorithm requires the remainder

r

when dividing by 5 to satisfy

0 \le r < 5

.

Anahtar Kavram

Modular cyclicity and negative remainder conversion

Tahmini Süre:2m 0s

Remainders and Units Digit Cyclicity

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