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Q: What number can be divided by 3 or 5 with a remainder of 1 and when divided by 7 and have no remainder?

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A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.A modulus is the remainder after a number has been divided into another. 3 divides into 10 with a remainder of 1. So 1 is the modulus of 10 divided by 3.

58

59

29

61,121,181,241,301,361,421,481,541,601,661,721,781,841,901,961,1021,1081,1141,1201, ect

solve it with a calculater

57

3 divided by 2 has a remainder of 1. Which is 1 less than 2.

Since the remainder is 0 when the numbers are divided by 3, then that number is a multiples of 3. For example, 45/3 = 15 remainder 0 45/4= 11 remainder 1 45/7 = 6 remainder 3

103

4

No. Add the digits of the dividend and if that is divisible by 3 then the original number is divisible by 3; if not, its remainder when divided by 3 gives the remainder when the original number is divided by 3: 1 + 2 + 1 = 4 which gives a remainder of 1 when divided by 3, so 121 divided by 3 gives a remainder of 1. (121 = 40 x 3 + 1)

24

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.

8

85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11

179 works until divided by 7 the remainder is 6. 2519 works till 10....

The answer is 58. 58/5 = 11 +3 58/4 = 14 +2 58/3 = 19 +1

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

16 and 31

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.

It will be 3.

87 is one of infinitely many numbers that meet these requirements.

365

The number is 119. 119/2= 64 R1, 119/3 = 39 R2, 119/4 = 29 R 3, and finally 119/5 = 23 R4. Hope this helps!