Q: What number is more than 10 less than 60 remainder of 1 when divided by 6 but remainder of 2 when divided by 5?

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The largest [integer] remainder is 10. If the remainder was any more you would get one (or more) lots of 11.

The LCM of 2, 3, 4 and 5 is 60. Since you need a remainder of 1 just add 1. So the answer is 61. Or any number that is 1 more than a multiple of 60.

An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. There are more odd products.

Restate the question to, "What multiple of 7 is one more than a multiple of 15?" The answer is 91.

The largest remainder would be 8, because if it were 9 you could divide the number once more. The largest remainder you can have is always one less than what you're dividing by. So if you're dividing by 10, your largest remainder is 9. If you're dividing by 100, it's 99. And so on.

An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. There s the same amount of each.

If the remainder is greater than the divisor then you can divide it once more and get one more whole number and then have less remainders.

58, 118, 178, ... That is all numbers that are 60n - 2 where n = 1, 2, 3, ... It has a remainder of 3 when divided by 5, means the last digit must be 3 or 8. It also has a remainder of 2 when divided by 4, means that the number must be even, so the last digit must be 8. It also has a remainder of 1 when divided by 3, means it must be 1 more than a multiple of 3 that ends in 7. So it must be 28, 58, 88, 118, ..., (each number 30 more than the last) but 28, 88, ... are divisible by 4, so only 58, 118, ... (each number 60 more than the last) need be considered. It also has a remainder of 4 when divided by 6, means it must be 4 more than a multiple of 6 that ends in 4. So it must be 58, 118, ...

11, 31, 51 and more at intervals of 20

you need to tell the number that you divided by. no one is able to figure it out until you give more information.

Since 17 is a factor of 119, the remiander will be the same modulo 17. Thus, the remainder, on division by 17 would be 19 except that 19 contains one more 17 - leaving a final remainder of 2.

5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.

A remainder can be any non-negative number that is less than the divisor. If the remainder is bigger than the divisor, the divisor can go into it another one (or more) times until the remainder is brought into that range.

27.2222

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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.

The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.

Not possible. Because if you do the number 21, you would say 14r7 is the answer right? WRONG. You can divide that 7 once more :D.

A common multiple of any two numbers, such as 11 and 18, is a number into which each of two or more number can be divided with zero remainder.

more info: An unknown polynomial f(x) of degree million yields a remainder of 1 when divided by x - 1, a remainder of 3 when divided by x - 3, a remainder of 21 when divided by x - 5. Find the remainder when f(x) is divided by (x - 1)(x - 3)(x - 5).

3 is the remainder when 23 is divided by 10. 2 x 10 = 20, and 23 is 3 more than 20.

There's no general rule or pattern. (11/5) divided by (33/5) = 1/3 (less than 1) (41/5) divided by (24/5) = 11/2 (greater than 1) Just as always in division . . . -- If you have (smaller number) divided by (bigger number), the quotient is less than 1. -- If you have (bigger number) divided by (smaller number), the quotient is more than 1.

Well, we know that it's one more than a multiple of 5 and one less that a multiple of 7. Also, it's even, because it has an even remainder when divided by an even number. So far, we know:It ends in 1 or 6 (remainder 1 when divided by 5)It's one less than a multiple of 7, which must therefore end in a 2 or a 7It ends with an even number, because it's evenTherefore it ends with a 6Great so far.Since it's one less than a multiple of 7, it has to also be one less than a multiple of 11, or 21, or 31 etc. because only numbers ending in a 1 will give 7 as a last digit when multiplied by 7.Now that we know it's either 7*11 - 1, or 7*21 - 1, or 7*31 - 1..., we start testing.7*11 - 1 = 7676/7 = 11r676/6 = 12r476/5 = 15r1All the criteria are met on the first try, therefore the number is 76.That was fun.

It might help to think of a division (with remainder) as "evenly distributing" some items - for example, give the same number of apples to each person. The "remainder" is whatever is LESS than the number of people (the divisor), so you can't continue distributing one more apple FOR EACH PERSON. If the remaining apples is greater than the number of people, or equal to them, you can distribute one more for each.

171.6