Question

Difficulty: HardRemainders and Units Digit Cyclicity

What is the remainder when the expression 745322+8157^{45} \cdot 3^{22} + 8^{15} is divided by 1010?

Answer: 5

Answer

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: 74571=7(mod10)7^{45} \equiv 7^1 = 7 \pmod{10}, 32232=9(mod10)3^{22} \equiv 3^2 = 9 \pmod{10}, and 81583=2(mod10)8^{15} \equiv 8^3 = 2 \pmod{10}. The expression simplifies to (7×9)+2=63+2=65(7 \times 9) + 2 = 63 + 2 = 65, which has a units digit of 5. Therefore, the remainder when divided by 10 is 5.

Step-by-Step Solution

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 NN can be expressed as 10k+r10k + r, where rr is the units digit and the remainder when NN is divided by 10.
2
Determine the units digit of 7457^{45}.
7457(mod10)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 7457^{45} has the same units digit as 71=77^1 = 7.
3
Determine the units digit of 3223^{22}.
3229(mod10)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 3223^{22} has the same units digit as 32=93^2 = 9.
4
Determine the units digit of the product 7453227^{45} \cdot 3^{22}.
7453223(mod10)7^{45} \cdot 3^{22} \equiv 3 \pmod{10}.
The product of the units digits is 7×9=637 \times 9 = 63, which has a units digit of 3.
5
Determine the units digit of 8158^{15}.
8152(mod10)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 8158^{15} has the same units digit as 83=5128^3 = 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=53 + 2 = 5.

Key Concept

Units Digit Cyclicity and Modular Arithmetic
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