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Difficulty: Very hardRemainders and Units Digit Cyclicity

Let n=1325−725n = 13^{25} - 7^{25}. When nn is divided by 1010, the remainder is R1R_1, where 0≤R1<100 \le R_1 < 10. When −n-n is divided by 77, the remainder is R2R_2, where 0≤R2<70 \le R_2 < 7. What is the value of R1+R2R_1 + R_2?

  1. A
    5
  2. 7Answer
  3. C
    10
  4. D
    12
  5. E
    2

Answer

The value of R1+R2R_1 + R_2 is 7.
To find R1R_1, we evaluate n=1325−725(mod10)n = 13^{25} - 7^{25} \pmod{10}. Both 3 and 7 have units digit cyclicity of period 4. Because 25≡1(mod4)25 \equiv 1 \pmod 4, 1325≡31=3(mod10)13^{25} \equiv 3^1 = 3 \pmod{10} and 725≡71=7(mod10)7^{25} \equiv 7^1 = 7 \pmod{10}. Then n≡3−7=−4≡6(mod10)n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}, so R1=6R_1 = 6. To find R2R_2, we evaluate −n=725−1325(mod7)-n = 7^{25} - 13^{25} \pmod 7. Since 725≡0(mod7)7^{25} \equiv 0 \pmod 7 and 13≡−1(mod7)13 \equiv -1 \pmod 7, we have −n≡0−(−1)25=1(mod7)-n \equiv 0 - (-1)^{25} = 1 \pmod 7, so R2=1R_2 = 1. Summing them yields R1+R2=6+1=7R_1 + R_2 = 6 + 1 = 7.

Step-by-Step Solution

1
Determine R1R_1, the remainder when n=1325−725n = 13^{25} - 7^{25} is divided by 10.
R1=6R_1 = 6
Working modulo 10, 13≡3(mod10)13 \equiv 3 \pmod{10}, so 1325≡325(mod10)13^{25} \equiv 3^{25} \pmod{10}. Powers of 3 follow a units digit cycle of period 4 (3,9,7,13, 9, 7, 1). Since 25=4×6+125 = 4 \times 6 + 1, 325≡31=3(mod10)3^{25} \equiv 3^1 = 3 \pmod{10}. Similarly, powers of 7 follow a units digit cycle of period 4 (7,9,3,17, 9, 3, 1), so 725≡71=7(mod10)7^{25} \equiv 7^1 = 7 \pmod{10}. Thus, n≡3−7=−4≡6(mod10)n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}, yielding R1=6R_1 = 6.
2
Determine R2R_2, the remainder when −n=725−1325-n = 7^{25} - 13^{25} is divided by 7.
R2=1R_2 = 1
Working modulo 7, 725≡0(mod7)7^{25} \equiv 0 \pmod 7. Since 13≡−1(mod7)13 \equiv -1 \pmod 7, 1325≡(−1)25=−1(mod7)13^{25} \equiv (-1)^{25} = -1 \pmod 7. Substituting these gives −n=725−1325≡0−(−1)=1(mod7)-n = 7^{25} - 13^{25} \equiv 0 - (-1) = 1 \pmod 7. Since 0≤1<70 \le 1 < 7, R2=1R_2 = 1.
3
Calculate the sum R1+R2R_1 + R_2.
6+1=76 + 1 = 7
Adding the two non-negative remainders gives R1+R2=6+1=7R_1 + R_2 = 6 + 1 = 7.

Key Concept

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

Remainders and Units Digit Cyclicity

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