Find the least number which when divided by any one of 12, 15 or 10 leaves 5 as remainder
Answers
Step-by-step explanation:
Let the number be denoted by k.
Therefore,
k=7m,
k=5a+2,
k= 10b+2,
k=12c+2,
k=15d+2.
Adding all of them and substituting each in form of b, we get,
k=10b+2.
Taking it as a common for all,
10b+2=7m
Now there is gonna be 2 at units place of this number, therefore, only 7×k6 have 2 at units place.
Therefore k starts from 0,1…
7×k6 -2 must be divisble by 5,10,12,15.
Therefore LCM of these 120, but 122 is not divisible by 122, i.e it doesn't lie in the A.P where a=42, d= 70 and nth term= 70n-28.
Therefore the number k should satisfy both these.
Therefore, n= 3, and therefore k=182
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Answer:
55
Step-by-step explanation:
First write some multiples of 12n-5 .Also find for 15n-5 and 10n-5 . Then the least common number is the answer. The least number is 55