What will be 6th number in this series?

Question

An ascending series of numbers satisfied the following conditions

When divided by 3, 4, 5 and 6 the number leaves the remainder of 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

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 ?

Chinese remainder theorem? https://en.wikipedia.org/wiki/Chinese_remainder_theorem — mfl, Sep 24 '18 at 18:11
Possible duplicate of A number when divided by 2, 3, 4, 5, 6 leaves a remainder of 1 but it is divided by 7 completely. — Dietrich Burde, Sep 24 '18 at 18:12
Always be on the lookout for underspecified problems. This problem doesn't say the sequence includes all numbers of the given form, so there is no way to tell which number will be 6th in the sequence. — Carl Mummert, Sep 24 '18 at 18:45

score 2 · Answer 1 · answered Sep 24 '18 at 18:23

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!

cansomeonehelpmeout · Answer 2 · 2018-09-24T18:35:41.473

1

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

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.

edited Sep 24 '18 at 18:35

answered Sep 24 '18 at 18:13

cansomeonehelpmeout

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score 1 · Accepted Answer · answered Sep 24 '18 at 18:24

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.

What will be 6th number in this series?

3 Answers3