What is the smallest whole number which:?

Question

when divided by 3, gives a remainder of 1, when divided by 4 gives a remainder of 2, when divided by 5, gives a remainder of 3, and when divided by 6, givs a remainder of 4.

I bet no one can get that!

Answer 1

It has to be even because of the base 4- condition. It has to end in 8 because of the base-5 condition.

It is 58?

Answer 2

First, the condition that n≡1 mod 3 is redundant, because any number which is congruent to 4 mod 6 will be congruent to 1 mod 3. Now, assuming n is even, n≡2 mod 4 iff that n/2 ≡ 1 mod 2, and similarly we have that n≡4 mod 6 iff n/2≡2 mod 3. Finally, using that 2⁻¹ mod 5 = 3, we see that n/2 ≡ 3*3 ≡ 4 mod 5. Letting k=n/2, we see that:

k≡1 mod 2
k≡2 mod 3
k≡4 mod 5

And if k satisfies all these congruences, then 2k will satisfy the original problem. There is a unique solution for k mod 30, which can be found using the Chinese remainder theorem, or more simply by noting that:

k+1≡0 mod 2
k+1≡0 mod 3
k+1≡0 mod 5

So k+1≡0 mod 30, k≡29 mod 30

Therefore, the smallest whole number which satisfies the given congruences is 2*29 or 58.

Answer 3

Every whole number that satisfies these conditions is of the form:

x = 60*k - 2, where k is an arbitrary positive integer,
so the smallest number is for k = 1, that is x = 60 - 2 = 58. Indeed:
58 = 3*19 + 1
58 = 4*14 + 2
58 = 5*11 + 3
58 = 6*9 + 4

Find the formula I've posted yourself!

Answer 4

LCM (3,4,5,6) - 2 = 58