Question

Find the smallest number which when increased by $ 17 $ is exactly divisible by both $ 520 $ and $ 468 $ ?

Answer 1

Hint: In the given question, we are required to find the smallest number which when increased by $ 17 $ is exactly divisible by both $ 520 $ and $ 468 $ . So, we first try to find the smallest number divisible by both $ 520 $ and $ 468 $ by taking the least common multiple of both the numbers. Then, we find our required answer by subtracting $ 17 $ from the LCM of $ 520 $ and $ 468 $ as the number has to be increased by $ 17 $ to be the LCM of the give numbers.

Complete step-by-step answer:
So, we first find the LCM of the numbers $ 520 $ and $ 468 $ .
To find the least common multiple of $ 520 $ and $ 468 $ , first we find out the prime factors of both the numbers.
Prime factors of \[520\] $ = 2 \times 2 \times 2 \times 5 \times 13 $
 $ = {2^3} \times 5 \times 13 $
Prime factors of \[468\] $ = 2 \times 2 \times 3 \times 3 \times 13 $
 $ = {2^2} \times {3^2} \times 13 $
Now, Least common multiple is a product of common factors with highest power and all other non-common factors. We can see that $ {2^2} $ and $ 13 $ are the common factor of $ 520 $ and $ 468 $ .
Hence, least common multiple of $ 520 $ and $ 468 $ \[ = {2^3} \times {3^2} \times 5 \times 13\]
 $ = 4680 $
Hence, the least common multiple of $ 520 $ and $ 468 $ is $ 4680 $ .
Now, we know that we have to find the smallest number which when increased by $ 17 $ is exactly divisible by both $ 520 $ and $ 468 $ .
So, the smallest number which is divisible by both $ 520 $ and $ 468 $ is $ 4680 $ as the LCM of $ 520 $ and $ 468 $ .
So, to get our required answer, we have to subtract $ 17 $ from the LCM of $ 520 $ and $ 468 $ , that we found as $ 4680 $ .
So, the required answer is $ 4680 - 17 = 4663 $ .
So, the correct answer is “4663”.

Note: Least common multiple (LCM) has wide ranging applications in real world as well as in mathematical questions. Knowledge of the least common multiple is also used in addition and subtraction of fractions.