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Remainder of 2222^5555 + 5555^2222 mod 462 | IOQM

JEE Maths question with a full step-by-step solution.

Question
Let N=22225555+55552222N=2222^{5555}+5555^{2222}. Let rr be the remainder when NN is divided by 462462. Find the sum of the distinct prime factors of rr.
Solution
Answer: 21
Step 1: Factor the divisor: 462=2×3×7×11462=2\times3\times7\times11. We examine NN modulo 2,3,72,3,7 and 1111 separately, using the identities: for any nn, an−bna^n-b^n is divisible by a−ba-b; for odd nn, an+bna^n+b^n is divisible by a+ba+b; and for even nn, an−bna^n-b^n is divisible by a+ba+b. Step 2 (divisibility by 11): Since 2222=11×2022222=11\times202 and 5555=11×5055555=11\times505, both bases are multiples of 1111, so both 222255552222^{5555} and 555522225555^{2222} are multiples of 1111. Hence NN is divisible by 1111. Step 3 (divisibility by 2): Here 222255552222^{5555} is even and 555522225555^{2222} is odd, so N=even+oddN=\text{even}+\text{odd} is odd. Thus NN is not divisible by 22. Step 4 (divisibility by 3): Write
N=(22225555+15555)+(55552222−12222).N=\big(2222^{5555}+1^{5555}\big)+\big(5555^{2222}-1^{2222}\big).
 The first bracket is a multiple of 2222+1=2223=3×7412222+1=2223=3\times741 (sum of like odd powers), and the second is a multiple of 5555+1=5556=3×18525555+1=5556=3\times1852 (even power). Both are divisible by 33, so NN is divisible by 33. Step 5 (divisibility by 7): Write
N=(22225555−35555)+(55552222−42222)+(35555+42222).N=\big(2222^{5555}-3^{5555}\big)+\big(5555^{2222}-4^{2222}\big)+\big(3^{5555}+4^{2222}\big).
 The first bracket is a multiple of 2222−3=2219=7×3172222-3=2219=7\times317, and the second is a multiple of 5555−4=5551=7×7935555-4=5551=7\times793. For the third, since 35555=(35)1111=24311113^{5555}=(3^5)^{1111}=243^{1111} and 42222=(42)1111=1611114^{2222}=(4^2)^{1111}=16^{1111},
35555+42222=2431111+161111,3^{5555}+4^{2222}=243^{1111}+16^{1111},
 which (sum of like odd powers) is a multiple of 243+16=259=7×37243+16=259=7\times37. All three brackets are divisible by 77, so NN is divisible by 77. Step 6: Collecting the results, NN is divisible by 3×7×11=2313\times7\times11=231 but is odd. The only multiples of 231231 less than 462462 are 00 and 231231, of which just 231231 is odd. Therefore
r=N mod 462=231=3×7×11.r=N\bmod 462=231=3\times7\times11.
 Step 7: The distinct prime factors of r=231r=231 are 3,7,113,7,11, so their sum is
3+7+11=21.3+7+11=21.
 Answer: 21
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