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Any number greater than 199 will divide into both zero times and leave a remainder (of the original number); there is no upper limit to the numbers greater than 199.

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The largest number to divide into both and leave the SAME remainder is 56 (leaving a remainder of 31 in both cases).

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There are infinitely many such numbers. Consequently, there is none which is largest.

Q: What is the largest number which when divided into 143 and 199 will leave a remainder in each case?

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Any number that can be divided by 2 and leave no remainder - for example, 2, 4, 6, 8, 10, is an even number.

3 can be divided into any number, but if you mean a number that will leave no remainder than it's still no.

No answer is possible as any number divided by 9 must either be exactly divisible by 9 or leave a remainder less than 9.

The first five numbers which when divided by 5 leave a remainder of 4 are: 4 = 4/5 = 0 remainder 4 9 = 9/5 = 1 remainder 4 14 = 14/5 = 2 remainder 4 19 = 19/5 = 3 remainder 4 24 = 24/5 = 4 remainder 4 The pattern continues in this way.

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Related questions

There is no such number. You can have arbitrarily large numbers with this property - therefore there is no largest.

a number that can be divided and leave no remainder.

92

A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.

If the number is divided by 119, it cannot be a prime!

Any number that can be divided by 2 and leave no remainder - for example, 2, 4, 6, 8, 10, is an even number.

All multiples of 60.

24

The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9.

3 can be divided into any number, but if you mean a number that will leave no remainder than it's still no.

1x54 , 2x27 , 3x18 , 6x9

This is an extremely poor question. There are 28 pairs of numbers and, in less than 10 minutes, I could find reasons why 17 pair were different from the other numbers in the set. A bit more time and I am sure I could do the rest. So here they are: 53 and 72: leave a remainder of 15 when divided by 19. 53 and 77: leave a remainder of 5 when divided by 12. 53 and 82: leave a remainder of 24 when divided by 29. 53 and 87: leave a remainder of 2 when divided by 17. 53 and 95: leave a remainder of 11 when divided by 21. 53 and 97: leave a remainder of 9 when divided by 11. 67 and 95: leave a remainder of 11 when divided by 28. 67 and 87: leave a remainder of 7 when divided by 17. 67 and 97: leave a remainder of 7 when divided by 30. 72 and 82: leave a remainder of 0 when divided by 2 (are even). 72 and 87: leave a remainder of 0 when divided by 3 (multiples of 3). 72 and 95: leave a remainder of 3 when divided by 23. 72 and 97: leave a remainder of 22 when divided by 25. 77 and 95: leave a remainder of 5 when divided by 9. 77 and 97: leave a remainder of 17 when divided by 20. 82 and 95: leave a remainder of 4 when divided by 13. 87 and 95: leave a remainder of 7 when divided by 8.