How many four-digit squares differ by 1 from a multiple of 10?

Answer: This slightly trickier variation on a previous teaser seemed to knock some of our Brainiacs out of shape. But not the following expert readers, who correctly answered 27: **William A Alexander**, associate professor, department of chemistry, drones research fellow, FedEx Institute of Technology, University of Memphis, Memphis, Tennessee, USA; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **France Veillette**: chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **David Mann**, key account manager, SPC Rubber Compounding, UK; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria. Well done to all and everyone else who had a go.

*Solutions*

*There were some nice variations in approach to this question:*

*William A Alexander*

Can be done by inspection of a simple spreadsheet, but there is a pattern. Four-digit squares run from 32 (1024) to 99 (9801) - 68 total possibilities to examine. If you look at the ones-place of the squares, the pattern (starting at 30) is 0,1,4,9,6,5,6,9,4,1,0; the pattern goes back and forth between 0 and 5. This back and forth pattern continues for all perfect squares to infinity.

We're looking for anything with a ones-place digit of 1 or 9, which corresponds to 4 out of every ten choices. Thinking of 70 choices - 30 to 99 - yields 28 total squares, and we remove 1 (for 31) to get to a final answer of 27 total squares that differ from a multiple of ten by 1.

The pattern is the key and explaining it takes way longer than doing it!

*John Bowen*

Smallest number is 33, largest is 99

For any set of numbers n1, n2, .... n9 where n=10s, the only ones with squares differing from a multiple of 10 are n1, n3, n7, and n9 which have squares ending 1,9,9,1.

So there are 3 in the 30s, and 4 in 40s, 50s, 60s, 70s, 80s, and 90s, so 27 in total.

*Andrew Knox*

If the question is interpreted as “which squares (of integers) with 4 digits differ by one from multiples of 10” i.e. end with a 1 or a 9, then the answer is 27.

31 squared is 961 (too few digits)

32 squared is 1024 (okay)

99 squared is 9801 (okay)

100 squared is 10000 (too many)

So, in the 30 series only 3 work, the other 6 series (40 series through 90 series), 4 work, so the answer is 3 + (4 x 6) = 27.

*France Veillette*

There are 68 four-digit squares, from 322 to 992. For the four-digit square to differ by one from a multiple of 10, it has to end by 1 or 9, therefore the number squared has to end with 1, 3, 7 or 9. Between 32 and 99, there are 27 such numbers

*David Mann*

The four-digit range is from 1000 to 9999, so the roots of the square numbers must lie in the range of 32 to 100. To differ by 1 from these, the roots must end in 1,3,7 or 9.

So, the numbers that meet the criterion are; 33, 37, 39 41, 43, 47, 49 And so on until 99.

That makes 27 numbers in all.

*Michele Girardi*

The problem can be reformulated as "how many integers, squared, yield a number between 1000 and 9999, that end by 1 or 9?"

The target numbers must end by 1,3,7,9 and be between 32 and 99.

There are 4 numbers per decade and 7 decades, for a total of 28, but we must exclude 31 that is out of range, so the numbers are 27.

New teaser on Monday.