The Greatest Positive Integer k for Which 49ᵏ + 1 Is a Factor of the Sum 49¹²⁵ + 49¹²⁴ + ⋯ + 49² + 49 + 1 Is

JEE Main 2020 — 7 January, Morning Shift. Previous Year Question.

Problem

The greatest positive integer $k$, for which $49^k+1$ is a factor of the sum $49^{125}+49^{124}+\cdots+49^2+49+1$, is:

Key insight. The sum is a geometric series that collapses to $\dfrac{49^{126}-1}{48}$ — and $126=2\times63$ means $49^{126}-1$ is a genuine difference of squares, $\left(49^{63}\right)^2-1^2$, which factors into $(49^{63}-1)(49^{63}+1)$. That second factor is exactly the form the problem is asking about.

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Approach

Recognise the sum as a geometric series and write it in closed form. Rewrite $49^{126}-1$ as a difference of two squares by splitting the exponent $126=2\times63$, and factor it into $(49^{63}-1)(49^{63}+1)$. Show that dividing by $48=49-1$ leaves the first factor as a genuine integer (itself a shorter GP sum), confirming $49^{63}+1$ divides the original sum exactly — then argue this is the largest such $k$.

Solution

Step 1 — Write the sum as a closed-form geometric series

$$S = 1+49+49^2+\cdots+49^{125} = \frac{49^{126}-1}{49-1} = \frac{49^{126}-1}{48}$$

Step 2 — Factor 49¹²⁶ − 1 as a difference of squares

Since $126=2\times63$:

$$49^{126}-1 = \left(49^{63}\right)^2-1^2 = \left(49^{63}-1\right)\left(49^{63}+1\right)$$

Step 3 — Divide by 48 and show the quotient is an integer

$$S = \frac{\left(49^{63}-1\right)\left(49^{63}+1\right)}{48}$$

Since $\dfrac{49^{63}-1}{48}=\dfrac{49^{63}-1}{49-1}$ is itself the closed form of the geometric series $1+49+49^2+\cdots+49^{62}$ (a sum of $63$ integer terms), it is a genuine integer.

Step 4 — Confirm the factorisation

$$S = \left(1+49+49^2+\cdots+49^{62}\right)\times\left(49^{63}+1\right)$$

This shows $49^{63}+1$ divides $S$ exactly, confirming $k=63$ works.

Step 5 — Why 63 is the greatest such k

The exponent $126$ splits as a difference of squares only through its own factor of $2$ (i.e. $126=2\times63$) — there’s no larger $k$ for which $49^k+1$ divides evenly into $S$, since the natural factorisation of $49^{126}-1$ caps out at this midpoint split. (This matches the structure of the well-known identity: $a^n+b^n$ divides $a^{2n}-b^{2n}$ exactly when the larger exponent is twice the smaller, giving the maximal such factor here.)

Answer

Common mistakes

• Trying to work with the sum directly as 126 separate terms instead of recognising the closed-form GP sum first. Converting to $\dfrac{49^{126}-1}{48}$ immediately opens the door to the difference-of-squares factorisation — without it, the problem looks intractable.
• Forgetting to verify that $\dfrac{49^{63}-1}{48}$ is actually an integer. This step is what confirms $49^{63}+1$ is a genuine factor (not just an approximate divisor) of the original sum — skipping it leaves the argument incomplete.

Practise next

• Find the greatest positive integer $k$ for which $7^k+1$ is a factor of the sum $7^{63}+7^{62}+\cdots+7+1$, using the same difference-of-squares factorisation approach.