AMC 10 Step by Step

2020 AMC 10A Problem 24

P21-25~8 minPrint

Let nn be the least positive integer greater than 10001000 for which gcd⁡(63,n+120)=21andgcd⁡(n+63,120)=60.\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60. What is the sum of the digits of nn ?

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Key insight

Each gcd says a number divides and a bigger power does not. The dividing parts give n = 237 plus multiples of 420; test 1077, 1497, 1917.

Solution

Each gcd condition contains a "must divide" part and a "must not go further" part. Use the first parts to list candidates, then use the second parts to throw candidates out.

1. First condition. 63=3×3×763 = 3 \times 3 \times 7. For gcd⁡(63,n+120)\gcd(63, n + 120) to be exactly 21=3×721 = 3 \times 7, the number n+120n + 120 must be a multiple of 2121 but not a multiple of 99 (otherwise the gcd would be 6363).
2. Second condition. 120=8×15120 = 8 \times 15 and 60=4×1560 = 4 \times 15. For gcd⁡(n+63,120)\gcd(n + 63, 120) to be exactly 6060, the number n+63n + 63 must be a multiple of 6060 but not a multiple of 88 (otherwise the gcd would be 120120).
3. List numbers with n+63n + 63 a multiple of 6060. Then nn is 60−63=−360 - 63 = -3 plus a multiple of 6060, so n=57,117,177,237,…n = 57, 117, 177, 237, \ldots (each 6060 more than the last).
4. Which of these make n+120n + 120 a multiple of 2121? Check them: 57+120=17757 + 120 = 177 (no), 117+120=237117 + 120 = 237 (no), 177+120=297177 + 120 = 297 (no), 237+120=357=21×17237 + 120 = 357 = 21 \times 17 (yes). So n=237n = 237 works for both "divides" parts.
5. The pattern repeats every 420420. Adding 420420 keeps n+63n + 63 a multiple of 6060 and n+120n + 120 a multiple of 2121, since 420=60×7=21×20420 = 60 \times 7 = 21 \times 20. And 420420 is the smallest step that does both. So the candidates above 10001000 are 237+840=1077237 + 840 = 1077, then 14971497, then 19171917, and so on.
6. Test the "not further" parts.
- n=1077n = 1077: n+120=1197n + 120 = 1197. Its digits add to 1818, so it is a multiple of 99. The gcd would be 6363. Throw it out.
- n=1497n = 1497: n+120=1617n + 120 = 1617, digits add to 1515, fine. But n+63=1560=8×195n + 63 = 1560 = 8 \times 195, a multiple of 88. The gcd would be 120120. Throw it out.
- n=1917n = 1917: n+120=2037n + 120 = 2037, digits add to 1212, not a multiple of 99. Good. n+63=1980n + 63 = 1980, and 1980÷8=247.51980 \div 8 = 247.5, not a whole number. Good. So n=1917n = 1917.
7. The digits of 19171917 add to 1+9+1+7=181 + 9 + 1 + 7 = 18.

The answer is (C) 18\boxed{\textbf{(C)}\ 18}.

Why this works

"The gcd is exactly dd" means dd divides the number, and the extra prime factors that would make the gcd bigger do not. The "divides" parts combine into one repeating list of candidates; the "does not divide" parts are easiest to use as a final check on just a few candidates.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Translate each gcd. gcd⁡(63,n+120)=21\gcd(63, n + 120) = 21 with 63=32⋅763 = 3^2 \cdot 7 means 21∣n+12021 \mid n + 120 but 9∤n+1209 \nmid n + 120. gcd⁡(n+63,120)=60\gcd(n + 63, 120) = 60 with 120=23⋅15120 = 2^3 \cdot 15 means 60∣n+6360 \mid n + 63 but 8∤n+638 \nmid n + 63.

Divisibility parts: n≡−120≡6(mod21)n \equiv -120 \equiv 6 \pmod{21} and n≡−63≡57(mod60)n \equiv -63 \equiv 57 \pmod{60}. Write n=57+60kn = 57 + 60k; then 57+60k≡15+18k≡6(mod21)57 + 60k \equiv 15 + 18k \equiv 6 \pmod{21}, so 18k≡12(mod21)18k \equiv 12 \pmod{21}, i.e. 6k≡4(mod7)6k \equiv 4 \pmod 7, giving k≡3(mod7)k \equiv 3 \pmod 7. So n≡57+180=237(mod420)n \equiv 57 + 180 = 237 \pmod{420}.

Candidates above 10001000: 1077,1497,1917,…1077, 1497, 1917, \dots. Check the exclusions:

  • n=1077n = 1077: n+120=1197n + 120 = 1197, digit sum 1818, divisible by 99. Then gcd⁡(63,1197)=63\gcd(63, 1197) = 63. Rejected.
  • n=1497n = 1497: n+120=1617n + 120 = 1617 (digit sum 1515, fine), but n+63=1560=8⋅195n + 63 = 1560 = 8 \cdot 195, so gcd⁡(1560,120)=120\gcd(1560, 120) = 120. Rejected.
  • n=1917n = 1917: n+120=2037n + 120 = 2037, digit sum 1212, not divisible by 99; n+63=1980n + 63 = 1980, and 980=8⋅122.5980 = 8 \cdot 122.5 is not a multiple of 88. Both gcds are exactly 2121 and 6060. Accepted.

Digit sum of 19171917: 1+9+1+7=181 + 9 + 1 + 7 = 18.

The answer is (C) 18\boxed{\textbf{(C)}\ 18}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.