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Find the sum of all the three digit numbers which leave the remainder 3 when divided by 5 |
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Answer» 99090 Its ap have 103, 108 , 113,118..............198 yeh ap ayge okay Jyoti you ap is wrong AP=103,113,123........193An=a+(n-1)×d193=103+(n-1)×1090=(n-1)×109=n-110=nS10=10/2(a+l)S10=10/2(103+193)S10=5×296S10=1480 |
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