Question:
The least positive integral value of \( \lambda \) such that \( 10^{50}+(3 \times(4)^{52})+\lambda \) is divisible by 9 is :
The least positive integral value of \( \lambda \) such that \( 10^{50}+(3 \times(4)^{52})+\lambda \) is divisible by 9 is :
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Find the remainder of each term on division by 9 using \( 10=9+1 \) and \( 4^{52}=(3+1)^{52} \).
Updated On: Nov 12, 2026
- 5
- 3
- 7
- 1
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The Correct Option is A
Solution and Explanation
\( 10^{50}=(9+1)^{50} \) leaves remainder 1 when divided by 9.
\( 3\times 4^{52}=3\times(1+3)^{52}=3\left[1+52\cdot 3+\binom{52}{2}3^2+\dots\right] \). Every term after the first inside the bracket is a multiple of 3, so \( 3\times 4^{52}=3+9k \), leaving remainder 3.
So the expression leaves remainder \( 1+3+\lambda = 4+\lambda \) modulo 9.
For divisibility by 9 we need \( 4+\lambda\equiv 0 \pmod 9 \), so the least positive value is \( \lambda=5 \).
Answer: (1) 5
\( 3\times 4^{52}=3\times(1+3)^{52}=3\left[1+52\cdot 3+\binom{52}{2}3^2+\dots\right] \). Every term after the first inside the bracket is a multiple of 3, so \( 3\times 4^{52}=3+9k \), leaving remainder 3.
So the expression leaves remainder \( 1+3+\lambda = 4+\lambda \) modulo 9.
For divisibility by 9 we need \( 4+\lambda\equiv 0 \pmod 9 \), so the least positive value is \( \lambda=5 \).
Answer: (1) 5
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