What will be 6th number in this series?

Question

An ascending series of numbers satisfied the following conditions

1. When divided by 3, 4, 5 and 6 the number leaves the remainder of 2.
2. When divided by 11, The number leaves no reminder.

The 6th number in this series will be...

I want to know how will we solve this

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I tried LCM(3, 4, 5, 6) which is 60

As said that when divides reminder will be 2

So the number will be in the form of : 60x+2

and it is also said that it divides by 11 and get reminder 0 so the number will be in the form of : 11y

We can say

60x + 2 = 11y

Now what I have to do ?

Answer 1

Consider the 11 numbers mod $60\times 11 = 660$:

$$60x+2 \in \{60(11z+0)+2, 60(11z+1)+2, \ldots, 60(11z+10)+2\}$$

Figure out which ones are zero (mod 11)

This gives:

$$660z+2, 660z+62, 660z+122, \ldots, 660z+602$$

You get:

$$660z+242 = 11(60z+22)$$

$$660z+242 = 60(11z+4)+2$$

So, I believe the 6th number will be when $z=5$ (assuming a 0-indexed sequence). So, 3542 satisfies all of the modular congruences and is the 6th positive integer to do so.

Answer 2

$$60x+2=11y\\66x-6x+2=11y\\-6x=-2+11y-66x\\-2(-6x)=-2(-2+11y-66x)\\x+11x=4-22y+132x\\\dots\\x=4+11(11x-2y)$$

Therefore $x$ may be written $x=4+11k$, which means that $$\boxed{60x+2=60(4+11k)+2=242+11\cdot 60k}$$ where $k$ is any integer.

Answer 3

Since we know that "if a number is divided by 11, then it leaves no remainder", then we know that one of the numbers in ascending order has to be a multiple of 11. Indicated by 11n, but if a number is divided by 3, 4, 5, and 6, then it must have a remainder of 2. Making the smallest number 2. So lets start with a simple series, $$2, 5, 8, 11, 14, 17.$$ notice we have a 11 so we divide that by 11 and indicate that $$a_4 = 11n$$

now to continue the series, $$2, 5, 8, 0, 3, 6$$ our $$a_6 = 6$$

side note: if this isn't something you were hoping for, tell me if it needs improvements!