If the number were one more, it would be divisible by 3, 4 and 5.
So it would be a multiple of LCM(3, 4, 5) = 60
So the answer is 60k - 1 where k is any integer.
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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!
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179 works until divided by 7 the remainder is 6. 2519 works till 10....
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!
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
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solve it with a calculater
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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.