Challenge Level

We had lots of good solutions to this problem. Congratulations to Molly and Catherine (Mount School, York) who reasoned as follows:

The number N had to end in a zero, so that it could divide by 10, but not 100 as then it would divide by 50. Also, it had to be at least 882 so we multiplied 882 by 5 to give us a number ending in a zero, as 882 has a factor of 2 but no factor 5.

882 x 5 = 4410 and 4410 divides by 10, 90, 98 and 882 and is NOT divisible by 50, 270, 686 or 1764. It is also a factor of 9261000.

Well done also to the Key Stage 3 Maths Club, Strabane Grammar School, N. Ireland, to Claire, Rhona, Rachel, Chris, Claire, Colin and Adam of Madras College St Andrew's, to Jamie and James of Hethersett High School, Norwich and to Caroline of Ipswich High School.

The following solution, using prime factorisation, was sent by Chew of Tao Nan School in Singapore.

10 = 2 x 5

90 = 2 x 3 ^{2} x 5

98 = 2 x 7 ^{2}

882 = 2 x 7 ^{2} x 3 ^{2}

The number N is divisible by all the above 4 numbers so N is divisible by their LCM.

LCM = 2 x 3 ^{2} x 5 x 7 ^{2} = 4410

But N is not divisible by 50, 270, 686 and 1764. We must then look at their factors to make sure N is not divisible by any one of them :

50 = 2 x 5 ^{2}

270 = 2 x 3 ^{2} x 5

686 = 2 x 7 ^{3}

1764 = 2 ^{2} x 3 ^{2} x 7 ^{2}

N does not have 5 ^{2} , 3 ^{3} , 7 ^{3}
and 2 ^{2} as factors.

9261000 = 2 ^{3} x 3 ^{3} x 5 ^{3} x 7
^{3} = (2 x 3 ^{2} x 5 x 7 ^{2} ) x 2
^{2} x 3 x 5 ^{2} x 7

Other good solutions came in, rather too late to be incorporated, from Mark of King Edward VI Camp Hill School, Birmingham, Helen of Madras College and Lizzie, Helen, Kayleigh, Hannah and Bethany of Maidstone Girls' Grammar School.