## Teaser 2529

**From The Sunday Times, 13th March 2011** [link]

A letter to

The Timesconcerning the inflated costs of projects read:When I was a financial controller, I found that multiplying original cost estimates by π used to give an excellent indication of the final outcome.

Interestingly, I used the same process (using 22/7 as a good approximation for π). On one occasion, the original estimate was a whole number of pounds (less than £100,000), and this method for the probable final outcome gave a number of pounds consisting of exactly the same digits, but in the reverse order.

What was the original estimate?

When originally published the amount was given as “less than £10,000”, which was raised to “less than £100,000” in the following issue. But even with this change the puzzle is still open to interpretation.

[teaser2529]

## Jim Randell 8:15 am

on16 March 2021 Permalink |What the puzzle isn’t clear on is what happens to any fractional part of the new estimate. (Which there would always be if you were to multiply by π).

If we round down (simply ignoring any fractional part) in the result, then we

canget a solution to the original puzzle:And we get the same result if we round to the nearest pound.

If we round up:

Raising the limit to £100,000 then we get the following additional solutions:

Rounding to nearest:

Which we also get if we round up, along with:

So the only way we can get a unique solution with the revised upper limit is to assume that, before rounding, the new estimate was a whole number of pounds, and so no rounding was necessary. In this case only the pair £22407 → £70422 remain as a candidate solution, and this was the required answer.

This makes a manual solution easier (and a programmed solution a little faster), as we know the original estimate must be a multiple of 7.

The following Python program lets you play with various rounding methods.

Run:[ @replit ]Solution:The original estimate was: £22,407.If the estimates had been constrained to be between £100,000 and £1,000,000 then there is a single solution whichever rounding method is chosen:

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## Frits 1:04 pm

on17 March 2021 Permalink |LikeLike

## John Crabtree 1:40 pm

on18 March 2021 Permalink |X = abcde, Y = edcba where X * 22 = Y * 7

and so 2e = 7a mod 10, and so a = 2 and e = 7

By digital roots, the sum of the digits in X = 0 mod 3. X = Y = 0 mod 3

X = 0 mod 7. Y = 0 mod 11, and so X = 0 mod 11

And so X = 231*M, Y = 726*M, and M = 7 mod 10

Or X = 1617 + 2310*N and Y = 5082 + 7260*N

By inspection X cannot have 2, 3 or 4 digits

70,000 < Y < 80,000 and so N = 9 or 10, which is invalid by inspection.

N = 9 gives X = 22407 and Y = 70422, which is valid.

If X 6 digits, there are only 14 possible values of N, ie 96 to 109, to check.

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## Frits 12:12 pm

on19 March 2021 Permalink |@John,

Maybe you used the congruent sign in your analysis and it was replaced by equal signs.

If so you might try the enclosure of text as mentioned in Teaser 2756.

If not the analysis is wrong as 0 mod 11 equals 0.

Please elaborate as I am not an expert in modular arithmetic (I gather X must be a multiple of 3, 7 and 11?).

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## John Crabtree 4:16 pm

on19 March 2021 Permalink |@Frits

I do not recall the congruence sign from when I was first introduced to modulo arithmetic nearly 50 years ago. I have always used the equals sign (perhaps incorrectly) followed by mod n.

X = 0 mod 3 means that X is divisible by 3. And so, yes, X is divisible by 3, 7 and 11. As it also ends in 7 the solution space becomes very small. HTH

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## Hugh Casement 12:25 pm

on19 March 2021 Permalink |I meant to post this on St Patrick’s day, before most of the other comments appeared.

I’m sure that an integer solution is intended, so no rounding is necessary.

The original estimate must therefore be an integer multiple of 7.

The result is even, so the original estimate (with its digits in reverse order) must start with 2: any larger value would yield a result with more digits.

The result is a multiple of 11, so the original estimate (using the same digits) must be too.

Admittedly, that still means a lot of trial & error if we don’t use a computer.

22/7 is a lousy approximation to pi! Much better is 355/113, but I think that can work only if the original estimate is allowed to have a leading zero.

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