The Student Room Group

UKMT Putting A 1 At Beginning / End Of Number

I have a number that is less than one million. Putting a 1 after it
makes it three times as large as putting a 1 before it. What is my
number?

My workings:
IMG_20171203_173613623.jpg
My thought is that if something multiplied by 3 ends in a 1, it must start with a 7.
So I put a 7 in the result as well (because they're the same number) but now if I carry the 2 from 7 * 3 = 21 then I get 3*x + 2 = 7 and this has no integer solutions for x, what do I do now?
3x + 2 = 7 has no integer solutions but the number doesn't need to be 7, just end in a 7. So 3x + 2 = (some number ending in 7). Does that help?
Reply 2
Original post by Retsek
I have a number that is less than one million. Putting a 1 after it
makes it three times as large as putting a 1 before it. What is my
number?

My workings:
IMG_20171203_173613623.jpg
My thought is that if something multiplied by 3 ends in a 1, it must start with a 7.
So I put a 7 in the result as well (because they're the same number) but now if I carry the 2 from 7 * 3 = 21 then I get 3*x + 2 = 7 and this has no integer solutions for x, what do I do now?

Hint: If you had a three digit number xx and you put a one before it you would get 1000+x1000+x. Does this make sense?

What about if you put a 1 after a 3 digit number?

The number doesn't have to be 3 digits but there aren't too many possibilities and you could solve for each until you end up with an answer.

There may be a quicker way that I haven't thought of.
Reply 3
Original post by Retsek

My thought is that if something multiplied by 3 ends in a 1, it must start with a 7.
So I put a 7 in the result as well (because they're the same number) but now if I carry the 2 from 7 * 3 = 21 then I get 3*x + 2 = 7 and this has no integer solutions for x, what do I do now?

It looks like you should be able to solve it using your method after the help given above (I didn't see that you had posted working). Stick to your way before trying the alternative method that I gave.

Turns out your method is better than mine!
(edited 6 years ago)
Reply 4
Original post by uponthyhorse
3x + 2 = 7 has no integer solutions but the number doesn't need to be 7, just end in a 7. So 3x + 2 = (some number ending in 7). Does that help?


Helped immensely, cheers man! I got the answer as 42857 now
Original post by Retsek
x


Putting your problem into maths, I thought of the answer being the solution to 3(x+10n)=10x+13(x+10^n)=10x+1 where nZ+n6n \in \mathbb{Z^+}|n \leq 6.

So x=3×10n17\displaystyle x=\frac{3\times10^n-1}{7}. The thing on the RHS is 299...99n times2\underbrace{99...99}_\text{n times}. You keep doing long division into 77 until you reach n=5n=5 and you stop as you get an integer (you could check n=6n=6 doesn't give an integer if you like). And yeah you get the same answer.
Reply 6
Original post by I hate maths
Putting your problem into maths, I thought of the answer being the solution to 3(x+10n)=10x+13(x+10^n)=10x+1 where nZ+n6n \in \mathbb{Z^+}|n \leq 6.

So x=3×10n17\displaystyle x=\frac{3\times10^n-1}{7}. The thing on the RHS is 299...99n times2\underbrace{99...99}_\text{n times}. You keep doing long division into 77 until you reach n=5n=5 and you stop as you get an integer (you could check n=6n=6 doesn't give an integer if you like). And yeah you get the same answer.

This is the way I did it. It's pretty quick once you've generated the equation for e.g. a 3 digit number.
I'll be annoying and say "by inspection, 42857".

Slightly less annoying (and something that although rarely useful, is occasionally, worth knowing):

1/7=0.1˙42857˙1/7 = 0.\dot{1}4285\dot{7}

If you write 142857 twice, we can read off the first 6 multiples by reading 6 sequential digits from the appropriate place:

1143228654751428571^14^32^28^65^47^5142857 (superscripts are to show where to start, not to indicate powers or anything):

E.g. to find 3 x 142857, start from the number with a superscript 3 to get: 428571

Also, 7 * 142857 = 999999
Reply 8
Original post by DFranklin
I'll be annoying and say "by inspection, 42857".

Slightly less annoying (and something that although rarely useful, is occasionally, worth knowing):

1/7=0.1˙42857˙1/7 = 0.\dot{1}4285\dot{7}

If you write 142857 twice, we can read off the first 6 multiples by reading 6 sequential digits from the appropriate place:

1143228654751428571^14^32^28^65^47^5142857 (superscripts are to show where to start, not to indicate powers or anything):

E.g. to find 3 x 142857, start from the number with a superscript 3 to get: 428571

Also, 7 * 142857 = 999999

Nice. I should have recognised the sequence of digits 142857.
Original post by DFranklin
x


I recognised the sequence of digits but didn't know any tricks. Thanks for sharing.

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