find the greatest number of four digits which when divided by 11 , 12 , 15 and 28 leaves 5 remainder in each case
Answers
Answer:
What is the greatest number of four digits which when divided by 11, 21, 15 and 28 leaves 5, 15, 9 and 22 as remainders respectively?
We need the greatest 4 digit Number to give different remainders simultaneously with 4 different diviosrs.
Let’s analyze:
We need a remainder 5 when divided by 11
If the number is 11 and the divisor is 11 the remainder is 0; if the number is 12 then the remainder is 1, therefore the number has to be 5 more than 11 i. e. 16, in order for the remainder to be 5.
Let’s look at it slightly differently
If the number is 11 and the divisor is 11 the remainder is 0; if the number is 1 below 11 i. e. 10 then the remainder is 10, therefore the number has to be 6 below 11 i. e. 5, in order for the remainder to be 5.
Please note than 5 + 6 = 11 i. e. 6 = 11 - 5
Thus, the conclusion is, if there’s a perfect multiple, say m, of a certain divisor, say d, and the expected remainder is, say r, then then either the number has to be m+r or m-d+r, which is nothing but m - (d-r).
If the number is 100 and divisor is 10 and expected remainder is 4 then the number would be 100+4 = 104 or 100 - (10–4) = 94. Of course there are more possibilities but you would always end up similarly.
Coming back to our problem:
With 11, remainder of 5 is 11–5 = 6 below the multiple of 11
With 21, remainder of 15 is 21–15 = 6 below the multiple of 21
With 15, remainder of 9 is 15–9 = 6 below the multiple of 15
With 28, remainder of 22 is 28–22 = 6 below the multiple of 28
Suddenly the solution to the problem is very simple. We need to first find the LCM of 4 numbers 11, 21, 15 and 28. Then we need to check if there’s any integral multiple of the LCM that still is a 4 digit number. Finally subtract 6 from the largest multiple to arrive at the required number.
So, the LCM of 11, 21, 15 and 28 is 11*21*5*4, which is 4620. If we multiply 4620 with 2 we get 9240 and with 3, we get a number that’s a 5 digit number. Thus, 9240 is the largest 4 digit common multiple of11, 21, 15 and 28. We need to go 6 below 9240 and that number is 9234.
9234 = 839*11 + 5 — Remainder is 5
9234 = 439*21 + 15 — Remainder is 15
9234 = 615*15 + 9 — Remainder is 9
9234 = 329*28 + 22 — Remainder is 22
So, 9234 is the required answer.