Remainder When 19^200+23^200 Is Divided by 49

JEE Main 2023 — 1 February, Morning Shift. Previous Year Question.

Problem

The remainder when $19^{200}+23^{200}$ is divided by $49$ is ____.

Key insight. $19$ and $23$ are both close to $21=3\times7$ — a multiple of $7$ whose square ($441=9\times49$) is divisible by $49$. Writing $19=21-2$ and $23=21+2$ and expanding both with the binomial theorem makes almost every term divisible by $49$ automatically, since each carries a factor of $21^2$ or higher — except for one surviving term, which needs a second round of the same trick.

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Approach

Write $19=21-2$ and $23=21+2$, and expand both $200$th powers using the binomial theorem. Odd-indexed terms cancel between the two expansions, and even-indexed terms double up — but all of these except the very last one carry a factor of $21^2$ (divisible by $49$) and vanish modulo $49$. What’s left is a single power of $2$, which — conveniently — can itself be rewritten as a power of $8=1+7$ and expanded by the binomial theorem a second time to extract the final remainder.

Solution

Step 1 — Rewrite 19 and 23 relative to 21

$$19 = 21-2, \qquad 23 = 21+2$$

Step 2 — Expand both using the binomial theorem

$$19^{200}+23^{200} = (21-2)^{200}+(21+2)^{200} = \sum_{k=0}^{200}\binom{200}{k}21^{200-k}\big[(-2)^k+2^k\big]$$

Step 3 — Simplify using odd/even cancellation

For odd $k$: $(-2)^k+2^k=0$. For even $k$: $(-2)^k+2^k=2\cdot2^k$. So only even $k$ survive:

$$19^{200}+23^{200} = 2\sum_{k \text{ even}}\binom{200}{k}21^{200-k}\cdot2^k$$

Step 4 — Identify which terms survive modulo 49

For every even $k$ from $0$ to $198$, the exponent $200-k \geq 2$, so $21^{200-k}$ contains $21^2=441=9\times49$ as a factor — divisible by $49$. Only the term $k=200$ (where $21^{200-k}=21^0=1$) survives:

$$19^{200}+23^{200} \equiv 2\cdot\binom{200}{200}\cdot2^{200} \pmod{49} = 2\cdot2^{200} = 2^{201}$$

Step 5 — Rewrite 2²⁰¹ as a power of 8

Since $201=3\times67$:

$$2^{201} = \left(2^3\right)^{67} = 8^{67} = (1+7)^{67}$$

Step 6 — Expand using the binomial theorem again

$$(1+7)^{67} = \sum_{k=0}^{67}\binom{67}{k}7^k$$

For $k\geq2$, each term contains $7^2=49$, so it vanishes modulo $49$. Only $k=0$ and $k=1$ survive:

$$8^{67} \equiv \binom{67}{0} + \binom{67}{1}\cdot7 \pmod{49} = 1+67\times7 = 1+469 = 470$$

Step 7 — Reduce 470 modulo 49

$$470 = 9\times49+29 = 441+29$$

$$470 \equiv 29 \pmod{49}$$

Answer

Common mistakes

• Trying to find a repeating remainder pattern by repeatedly squaring 19 or 23 directly. Since neither number has a small order modulo 49, this approach never terminates in a clean cycle — the binomial-theorem route (relative to the nearby multiple of 7) is what actually works.
• Losing track of the extra factor of 2 from combining the two expansions’ matching terms. The surviving term is $2\times2^{200}=2^{201}$, not just $2^{200}$ — dropping this factor of $2$ leads to reducing the wrong power in the second binomial step.

Practise next

• Find the remainder when $17^{150}+25^{150}$ is divided by $49$, using the same two-stage binomial theorem approach (writing $17=21-4$, $25=21+4$).