the greatest numbers which divides 170 and 250, leaving remainder 2 and 5 reapectively is?
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Given that the numbers 170 and 250 leaving remainders 2 and 5 respectively.
The numbers that are exactly divisible are 170 - 2 = 168 and 250 - 5 = 245.
Prime factorization of 168 = 2 * 2 * 2 * 3 * 7
Prime factorization of 245 = 5 * 7 * 7
Therefore HCF(168,245) = 7.
Therefore The largest Which divides the 170 and 250 leaving remainder 2 and 5 = 7.
Hope this helps!
The numbers that are exactly divisible are 170 - 2 = 168 and 250 - 5 = 245.
Prime factorization of 168 = 2 * 2 * 2 * 3 * 7
Prime factorization of 245 = 5 * 7 * 7
Therefore HCF(168,245) = 7.
Therefore The largest Which divides the 170 and 250 leaving remainder 2 and 5 = 7.
Hope this helps!
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Answer:
Given that the numbers 170 and 250 leaving remainders 2 and 5 respectively.
The numbers that are exactly divisible are 170 - 2 = 168 and 250 - 5 = 245.
Prime factorization of 168 = 2 * 2 * 2 * 3 * 7
Prime factorization of 245 = 5 * 7 * 7
Therefore HCF(168,245) = 7.
Therefore The largest Which divides the 170 and 250 leaving remainder 2 and 5 = 7.
Step-by-step explanation:
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