The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
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List-I
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List-II
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(A) The least number that must be subtracted from 2025 to get a number exactly divisible by 17
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(II) 2
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(B) The least number that must be added to $1057^5$ to get a number exactly divisible by 23
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(IV) 1
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(C) Unit digit of $6^{15}-7^4$
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(I) 5
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(D) Find the product of any number and the 1st whole number
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(III) 0
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(A) Least number subtracted from 2025 to be divisible by 17
We divide $2025$ by $17$ to find the remainder:
- $2025 \div 17$:
- $17 \times 100 = 1700$
- $2025 - 1700 = 325$
- $17 \times 19 = 323$
- Remainder: $325 - 323 = 2$
- Since the remainder is $2$, we must subtract 2 from $2025$ to make it divisible.
- Match: (A) $\rightarrow$ (II)
(B) Least number added to 1057 to be divisible by 23
We divide $1057$ by $23$ to find the remainder:
- $1057 \div 23$:
- $23 \times 40 = 920$
- $1057 - 920 = 137$
- $23 \times 5 = 115$
- Remainder: $137 - 115 = 22$
- To find what to add, we subtract the remainder from the divisor: $23 - 22 = 1$.
- Match: (B) $\rightarrow$ (IV)
(C) Unit digit of $6^{15} - 7^4$
- Unit digit of $6^{15}$: Any power of $6$ always ends in $6$ (e.g., $6, 36, 216$). So, it is 6.
- Unit digit of $7^4$: $7^1=7, 7^2=9, 7^3=3, 7^4=1$. So, it is 1.
- Calculation: $6 - 1 = 5$.
- Match: (C) $\rightarrow$ (I)
(D) Product of any number and the 1st whole number
- The first whole number is 0.
- Any number multiplied by $0$ is always 0.
- Match: (D) $\rightarrow$ (III)
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