A number has remainder of 1 when divided by 2.It has a remainder of 2 when divided by 3.the remainders are 4,5 and 0 when it is divided by 5,6 and 7 respectively.Find the smallest possible value of this number,very urgent.
Asked by | 7th Apr, 2012, 11:52: AM
The number has remainder 1 when it is divided by 2.
This means it is an odd number.
The number has remainder 4, when divided by 5. A digit that is divisible by 5 ends in 0 or 5.
If we add 4 to this digit, we get the resulting digit either ending in 4 or 9. Since a digit ending in 4 becomes even, this means that the digit has to end in 9.
The number has remainder 0 when divided by 7, meaning that it is divisible by 7.
So the least digit divisible by 7 ending in 9 is 49.
Now , we need to check 49 if it satisfies all the given conditions.
49 divided by 2 has remainder 1
49 divided by 3 has remainder 1
So the answer is not 49, we need to find the next number in the table of 7 ending in 9. that is 70+49= 119. Now we need to verify all the given conditions
119 divided by 2 has remainder 1
119 divided by 3 has remainder 2
119 divided by 5 has remainder 4
119 divided by 6 has remainder 5
119 divided by 7 has remainder 0
So the least possible value for this number is 119
Answered by | 9th Apr, 2012, 06:47: PM
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