find d smallest no. which when increased by 17 is exactly divisible by both 520 and 468. pls answer fast
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Answer: the no is, 4663
Step-by-step explanation: Let the no be 'X'since it is smallest no divisible by both 520 & 468
therefore, X=HCF(520,468) -17
HCF(520,468)=4680[by prime factorisation ]
Since 4680 is the smallest no divisible by 520 and 468, which is increased by 17
therefore, X=HCF(520,468) -17 X=4680-17=4663
X=4663
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