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# Sunday Times Teaser 2691 – Hot X Buns

### by Victor Bryant

In the words EASTER, HOT and BUNS, I have consistently replaced different letters with different digits to produce numbers for which EASTER is exactly divisible by 7, 365, and the product HOT x BUNS.

What are HOT and BUNS in their numeric form?

3 Comments Leave one →
1. brian gladman permalink

• geoffrounce permalink

2. Peter Hayes permalink

Here it is manually in two parts.

Part one:

You can work this one out in your head (in the dark in bed). We have

EASTER = HOT x BUNS x p (dummy variable)

where 5, 7 and 73 divide EASTER evenly.

HOT x BUNS >= 200,000, therefore p < 5 Therefore 5 divides HOT or BUNS, which implies either T or S = 5, and R = 0. This then implies 73 divides EASTE. Every schoolboy knows that 10001 = 73 x 137, so we can cross out both of the E's. Therefore 73 divides AST, where either S or T = 5. 073, 146, 219, 292, 365 (aha!), 438, 511, 584, 657 (aha!!), 730, 803, 876, 949; that's all folks. So EASTER is either E365E0 OR E657E0. Part two: Divisibility by 7 (I'll skip this step here, but another easy head task) then gives EASTER = 536550 (obviously no good) or 265720 (which would make HOT x BUNS > 300,000 and thus no good) or 965790.

So EASTER = 965790 = HO7 x BUN5 and p is then obviously even, i.e. either 2 or 4. But as 965790 is not divisible by 4, p = 2.

HO7 x BUN5 = 482895 = 3 x 3 x 3 x 5 x 7 x 7 x 73, with {H, O, B, U, N} = {1, 2, 3, 4, 8}

Probably the easiest approach here on is to observe that no multiple of 73 will work for HO7;
it isn’t divisible by 5 obviously either, and being greater than 49, it must also be divisible by 3.

Modulo 3 then, we can see HO7 must be 387 (has 43 as a factor; no good), 837 (factor of 31; ibid), 237 (factor 79, no good)
327 (factor 109, no good), 417 (factor 13, no good) or, ahem, 147, which = 3 x 7 x 7.

So HOT = 147, BUNS = 3285, and to all A GOOD NIGHT!

Happy Easter!!!