Bright Way /Xurree Ifaa/

Find the smallest number which, on being added 23 to it, is exactly divisible by 32, 36, 48 and 96. **********************************************SOLUTION:
The smallest number which, on being added 23 to it means the smallest number x plus 23.
Then x + 23 is exactly divisible by 32, 36, 48 and 96. But, since x is the smallest number, x+ 23 is the least common multiple of 32, 36, 48, and 96. (i.e, x + 23= LCM(32, 36, 48, 96)).
Now to find the value of x, we have to know first LCM(32, 36, 48, 96).............. Use the prime factorization method to easily find LCM(32, 36, 48, 96)
prime factorization of 32 = 2^5. prime factorization 36 = 2^2×3^2
prime factorization 48 = 2^4×3. prime factorization 96 = 2^5×3
Then LCM = The common prime factor with greatest power times prime factors that are not common for all.
common prime factor with greatest power = 2^5;
prime factors that are not common for all = 3^2;
LCM(32, 36, 48, 96) =2^5×3^2 = 2×2×2×2×2×3×3 = 288
Then, x+23= LCM(32, 36, 48, 96)
x+23= 288 (subtract 23 from both sides to leave x alone)
x+23-23= 288 -23
x = 265

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