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What would happen if you did it the other way? Keep subtracting by 3 until it is divisible by 5? I am assuming that would be the method of solution if you needed a multiple of 5 for "5"'s and a multiple of 3 for "3"'s?
But look at the problem carefully my dear friend we have to find the largest number, But if we do it other way around than no. if 3s will be more thn no. of 5s hence it wont be the largest n. of that type.
Thanks a lot :-)