what is the least no which when divided by tye numbers 3 5 6 8 10 and 12 leaves in each case 2 but which when divided by 13 leaves no rrmainder?
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Answered by
1
Answer:
The number is 962
Step-by-step explanation:
Lcm of ( 3,5,6,8,10,12) * K + 2
= 120K+2
120K +2 /13
=117K+3K +2/13
The least value of K for which this holds true is 8 (3*8+2 = 26)
Therefore the least such number is 120*8+2 = 962
Answered by
0
Answer:
above answer is correct ....
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