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 s the same amount of each.
The remainder is 0.If A has a remainder of 1 when divided by 3, then A = 3m + 1 for some integer mIf B has a remainder of 2 when divided by 3, then B = 3n + 1 for some integer n→ A + B = (3m + 1) + (3n + 2)= 3m + 3n + 1 + 2= 3m + 3n + 3= 3(m + n + 1)= 3k where k = m + n + 1 and is an integer→ A + B = 3k + 0→ remainder when A + B divided by 3 is 0-------------------------------------------------------------------------From this, you may be able to see that:if A when divided by C has remainder Ra; andif B when divided by C has remainder Rb; then(A + B) divided by C will have remainder equal to the remainder of (Ra + Rb) divided by C
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. 157 is an odd number.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. 16 is an even number.
It is an integer which, when divided by 2, leaves a remainder of 1.
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
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59
solve it with a calculater
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17
Put the remainder over the number you originally divided by. 15 divided by 7 = 2, remainder 1 = 2 and 1/7
No number can satisfy these conditions: To have a remainder of 1 when divided by 6, the number must be odd (as all multiples of 6 are even and an even number plus 1 is odd) To have a remainder of 2 when divided by 8, the number must be even (as all multiples of 8 are even and an even number plus 2 is even) No number is both odd and even. → No number exists that has a remainder of 1 when divided by 6, and 2 when divided by 8.
58
It is not possible, because the number 4 is divisible by 2, and it's remainder is divisible by 2 also, so whatever number works for the "4 with a remainder of 2", will never work for "2 with a remainder of 1.
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.