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JIPMAT 2023

Number System
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Miscellaneous

Medium

Amicable numbers are a pair of distinct natural numbers (a, b) such that the sum of the proper divisors of a equals b and the sum of the proper divisors of b equals a. Given that (220, y) is a pair of amicable numbers, y equals:

Correct Option: 2
1) First, let's find all proper divisors of 220:\newline * Proper divisors are all positive divisors excluding the number itself\newline * Factors of 220 = 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220\newline * Proper divisors of 220 = 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110\newline * Sum of proper divisors of 220 = 284
2) Let's check each option:
If y = 225:\newline * Proper divisors of 225 = 1, 3, 5, 9, 15, 25, 45, 75\newline * Sum = 178\newline * Since 178 ≠ 220, 225 is not correct
If y = 284:\newline * Proper divisors of 284 = 1, 2, 4, 71, 142\newline * Sum = 220\newline * Since sum of proper divisors of 220 = 284 AND\newline * Sum of proper divisors of 284 = 220\newline * This makes (220, 284) an amicable pair!
No need to check 404 since we found our answer.
Therefore, y = 284.

Related questions:

JIPMAT 2022

Given below are two statements:
Statement I : The sum of exponents of prime factors in the prime factorization of 392 is 5.
Statement II : The decimal representation 1323×5\frac{13}{2^3 \times 5} will terminate after 2 decimal places.
In the light of the above statements, choose the correct answer from the options given below:

JIPMAT 2023

Given below are two statements:

Statement (I): The difference, the sum, and the product of two numbers are in the ratio 1: 5: 12. The product of the two numbers is 18.

Statement (II): The last digit in the decimal representation of 291 is 6.

In light of the above statements, choose the most appropriate answer from the options given below.

JIPMAT 2023

The expression 2x3 + ax2 + bx + 3, where a and b are constants, has a factor of x-1 and leaves a remainder of 15 when divided by x + 2. Then, (a, b) =

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