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Answers
♧QUESTION♧
find the smallest number which when divided by 24 36 and 54 gives a remainder of 5 each time
☆ANSWER☆
Let us assume the no. is of the format 24m+5, 36l+5 and 54n+5.Since all these formats represent the same number we can equate them.Thus 4m+5=6l+5=>4m=6lSimilarly we get4m=9n, and 6l=9n.So, 4m=6l=9n.Here on we guess the values of m, n and l such that 4m=6l=9n by taking LCM of 4, 6 and 9.We clearly see that m=9, n=6, l=4.So the answer is 24(9)+5
ANSWER=221
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Which is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?
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The smallest such number has to be 5! But if we’re not considering that, then here’s the solution:
The numbers which leave remainder 5 when divided by 24 are: 29, 53,.. [(a multiple of 24)+5].
Similarly, the numbers which leave remainder 5 when divided by 36 are: 41, 77,... [(a multiple of 36)+5]
And finally, the numbers which leave remainder 5 when divided by 54 are: 59, 113,.. [(a multiple of 54)+5].
Here we want one number that does all three! So it has to be a number that is =[(multiple of 24 and 36 and 54)+5] = (A common multiple)+5.
But because we want the least such number, we take least common multiple of 24, 36 and 54 and then add 5.
Let’s now work on finding the lcm of these three numbers.
24=2*2*2*3
36=2*2*3*3
54=2*3*3*3
Now the lcm is the number obtained by taking the highest available power of each prime factor.
So the lcm is = (2^3) * (3^3) =216.[The Highest power available is 3 for 2(in 24), 3 for (in 54)].
Answer = 216+5=221.