find the smallest number which leave remainder 4 in each case when divided by 22 and 26?
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Prime factorizations first.
12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
27 = 3 x 3 x 3
LCM = 2 x 2 x 3 x 3 x 3 x 5 = 540
Since all of the remainders are 4 less than the divisors, the number will be 540 - 4 = 536
sonali9898:
srry this is not a answer
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