## Listener Crossword 4114: Solution Notes Back to 2010 menu

## Three-Square by Elap

### Puzzle explanation

When the corner cells are filled correctly, all rows and columns are triangular numbers. The formula for the nth triangular number is T = n(n + 1)/2, so n can be found when most of the digits of T are known, as the next integer below √(2T).

NB: A primitive triangle is one where there is no factor common to all three sides, meaning that it can’t be scaled down to a similar triangle with integer sides. Two of the three sides may have a common factor, however.

### Puzzle solution process

L2 and T2 are both less than 100, so L2 * T2 is less than 10000. From the clue 33333 = L2 * T2 + r5, r5 must then start with 2 or 3. From 333333 = E3 * P2 * j2, j2 is odd, so the first digit of P and r is odd; thus r5 and P2 start with 3, and j2 ends with the same 3.

From 3333333 = P2 * X6 + W3, P2 can’t be greater than 33 because X6 has six digits, so P2 is 31 or 33. From 333333 = E3 * P2 * j2, P2 has to be 33, because 31 isn’t a factor of 333333. This leads to j2 = 13 (the only two-digit factor of 333333 / 33 that ends in 3) and E3 = 777.

X6 is now between (3333333 − 999) / 33 and (3333333 − 100) / 33, so it’s in the range 100980 to 101007; therefore X6 starts with 10 and X3 is 100 or 101. From X3 * X3 = R2 + g4, where R2 and g4 have the same last digit, X3 must be even, so X3 = 100 and X6 is 1009__. R2 + g4 is now 10000. Their shared final digit can’t be 0 because it’s the first digit of S and t, so it must be 5. Thus g4 is 99_5.

In the right-angled triangle E3, E3 − M3, V2 * X2 + Q3, E3 = 777 can’t be the side corresponding to 2pq (using the information from the preamble), so it must be p2 − q2 = (p + q)(p − q) = 777. The other side, 2pq = E3 − M3, ends in 4 because E3 = 777 and M3 ends in 3. The hypotenuse, p2 + q2 = V2 * X2 + Q3, ends in 5 because X2 = 10 and Q3 ends in 5. The only possibilities for factorising 777 = 3 * 7 * 37 with (p − q) less than (p + q) are:
 p − q p + q p q 3 259 131 128 7 111 59 52 21 37 29 8
Of these, only the last gives a value of 2pq ending in 4, namely 2 * 29 * 8 = 464, so E3 − M3 = 464, giving M3 = 313, and the hypotenuse is 905.

Using h2 * h2 = S3 + i2, we know S3 starts with 5, so S3 + i2 is in the range from 500 + 10 = 510 to 599 + 99 = 698. This puts h2 in the range 23 to 26, making the first digit 2. Therefore D2 = 27. In the right-angled triangle D2, s3, q3, D2 = 27 must be the p2 − q2 side (being odd), and s3 = 2pq and q3 = p2 + q2 both start with 3. The only possibilities for factorising 27 are:
 p − q p + q p q 1 27 14 13 3 9 6 3
Only the first of these has p and q relatively prime and gives three-digit values for the other sides, namely s3 = 364 and q3 = 365.

For p3 * p3 = f3 + w5 we have _71 + 5____, which is in the range 171 + 50000 = 50171 to 971 + 59999 = 60970, so p3 starts with 2, which makes L2 = 23.

For 3333333 = P2 * X6 + W3, we have 3333333 = 33 * 1009__ + __4, so X6 must end with a 3 and is therefore either 100983 with W3 = 894, or 100993 with W3 = 564. For 33333 = L2 * T2 + r5 we have 33333 = 23 * 6_ + 3____; r5 therefore is from 33333 − 23 * 69 = 31746 to 33333 − 23 * 60 = 31953. Since we know that the middle digit of W3 is either 6 or 9, r3 must be 319__, which makes W3 = 894 and X6 = 100983.

For s3 * s3 = j6 + v3 with s3 = 364 and j6 = 13_9__, j6 is from 132496 − 999 = 131497 to 132496 − 100 = 132396, so j6 must start with 1319. Then from 333 = B3 + N2 + j3 with N2 = 13 and j3 = 131, we have B3 = 189.

For p3 * p3 = f3 + w5 we now have _71 + 58___, so p32 is in the range 171 + 58000 = 58171 to 971 + 58999 = 59970, which gives p3 = 242 to 244, so the second digit of p3 is 4.

For the right-angled triangle G2, Q2, k2, we know k2 is _3 = p2 + q2. All perfect squares end with 0, 1, 4, 5, 6 or 9; a sum of two squares can end in 3 only if one of them ends with 4 and the other ends with 9. The only possibility for a two-digit sum is 64 + 9, so we have p = 8 and q = 3, giving k2 = 73 and the other sides (in some order) as p2 − q2 = 55 and 2pq = 48. From e3 * e3 = F5 + Q3 we know that e3 is even (because F5 ends in 1 and Q3 ends in 5), so G2 = 48 and Q2 = 55. Then R2 = 55 and g4 = 9945, from X3 * X3 = R2 + g4.

For the hypotenuse X2 * V2 + Q3 we now have 10 * V2 + 555 = 905, so V2 = 35 and its first digit completes p3 = 243. From e3 * e3 = Q3 + F5 with e3 = 124 and Q3 = 555, we now get F5 = 14821. This makes b4 = _878, and we have W2 = 89, so from W2 * W2 = 7921 = b4 + u2 we get b4 = 7878 and u2 = 43.

For p3 * p3 = f3 + w5 we have 2432 = 59049 = _71 + 58___, so w5 is 58_78. For h2 * h2 = S3 + i2 we have 242 = 576 = 544 + i2, so i2 = 32. For i2 * i2 = C3 + Q2 we now have 322 = 1024 = __9 + 55, so C3 = 969. For Z2 * Z2 = G2 + u4 we have Z22 = 48 + 430_, so Z2 = 66 and u4 = 4308.

For the scalene triangle H2 + p3, h3 + n3, n3, m5, the sides are 17 + 243, 242 + 25_ and 25_, ie, 260, 492 + x and 250 + x, where x is the last digit of n3. Since m5, the area of the triangle, is the square root of s(s − a)(s − b)(s − c), there must be an even number of each prime factor in this value. The semi-perimeter, s, is (260 + 492 + x + 250 + x) / 2 = 501 + x. One of the values in the product is (s − n3), which is (501 + x) − (250 + x) = 251, a prime. There must therefore be another 251 factor somewhere, ie, in one of (501 + x), (501 + x) − 260 = 241 + x, or (501 + x) − (492 + x) = 9. The last of these, 9, obviously doesn’t have a factor of 251 and, since x is a single digit, 241 + x can’t either. Therefore 501 + x must be divisible by 251, giving x = 1. So, n3 = 251 and the area is m5 = 16566.

For s3 * s3 = j6 + v3 we now have 3642 = 132496 = 1319__ + 566, giving j6 = 131930. For 33333 = L2 * T2 + r5 we then have 33333 = 23 * T2 + 31930, so T2 = 61.

For the right-angled triangle a2 + r2, e2 + m2, t2, the sides are a2 + 31, 12 + 16 = 28 and a hypotenuse of 5_. If 28 is the p2 − q2 side, the only way to factorise 28 = (p + q)(p − q) so that p and q are positive integers is with p + q = 14, p − q = 2, p = 8 and q = 6. This gives a hypotenuse of 100, which is wrong. So 28 is the 2pq side, which makes p = 7 and q = 2, resulting in a triangle with sides 45, 28 and 53, and making a2 = 14 and t2 = 53.

For 3333 = B3 * a2 + Y3, we 3333 = have 189 * 14 + Y3, so Y3 = 687. For p3 * p3 = f3 + w5, we have 59049 = _71 + 58878, so f3 = 171.

For the scalene triangle U3, h2 + k2, j3 + v3, N4 + d5, the sides are 65_, 24 + 73 = 97 and 131 + 566 = 697. Using x for the third digit of U3, the semi-perimeter, s, is 722 + x / 2, so x must be even. If the sides are a, b and c, we have five possibilities:
 x a b c s s − a s − b s − c 0 650 97 697 722 72 625 25 2 652 97 697 723 71 626 26 4 654 97 697 724 70 627 27 6 656 97 697 725 69 628 28 8 658 97 697 726 68 629 29
We need an even number of each prime factor in s(s − a)(s − b)(s − c). For x = 2, there’s a factor of 71 in s − a only; for x = 4, there’s a factor of 5 in s − a only; for x = 6, there’s a factor of 23 in s − a only; and for x = 8, there’s a factor of 29 in s − c only. That leaves x = 0, where all the prime factors can be seen to pair off: (2 * 19 * 19)(2 * 2 * 2 * 3 * 3)(5 * 5 * 5 * 5)(5 * 5), and so we have U3 = 650. The area is therefore 2 * 2 * 3 * 5 * 5 * 5 * 19 = 28500 = N4 + d5, so d5 = 27164.

For the scalene triangle J3 + x2, W2 + m2, p3 + x2, a5, the sides are 246 + 40 + x = 286 + x, 89 + 16 = 105 and 243 + 40 + x = 283 + x, where x is the second digit of x2. Once again, looking at the semi-perimeter, we have s = 337 + x. Two of the terms of s(s − a)(s − b)(s − c) are therefore (337 + x) − (286 + x) = 51 and (337 + x) − (283 + x) = 54. The 51 term has a prime factor of 17, so there must be another factor of 17 in s = 337 + x or s � 105 = 232 + x. Since x is in the range 0 to 9, it must be 3 or 6:
 x a b c s s − a s − b s − c 3 105 289 286 340 235 51 54 6 105 292 289 343 238 51 54
For x = 3, the factor of 47 in 235 is unmatched, so x = 6, making x2 = 46. The prime factors are (7 * 7 * 7)(2 * 7 * 17)(3 * 17)(2 * 3 * 3 * 3) and the area is 2 * 3 * 3 * 7 * 7 * 17 = 14994 = a5.

For the scalene triangle A3 + w2, B2 + c3, C3, K3 + q5, the sides are _42 + 58 = 100x + 100, 18 + _63 = 100x + 81 and 969, where x is the first digit of A3 and c3. The semi-perimeter, s, is (100x + 100 + 100x + 81 + 969) / 2 = 100x + 575, making s − a = 475, s − b = 494 and s − c = 100x − 394. The area is K3 + q5 = 462 + 36588 = 37050, and so we have 370502 = s(s − a)(s − b)(s − c) = (100x + 575) * 475 * 494 * (100x − 394), which reduces to (100x + 575)(100x − 394) = 5850. The first term is at least 575, so the second term can’t be more than 10 to make a product of 5850. Therefore x is 4, giving A3 = 442 and c3 = 463.

For the scalene triangle G3 + L2, N3 + z2, y3, b4, the sides are 482 + 23 = 505, 133 + 7_ = 203 + x and 30_ = 300 + y, where x is the second digit of z2 and y is the third digit of y3. The area is 7878, which is 2 * 3 * 13 * 101. We need an even number of the 101 factor in the product s(s − a)(s − b)(s − c). The semi-perimeter, s, is (505 + 203 + x + 300 + y) / 2 = 504 + x / 2 + y / 2, and so the product is (504 + x / 2 + y / 2)(x / 2 + y / 2 − 1)(301 − x / 2 + y / 2)(204 + x / 2 − y / 2). Two of these terms must have a factor of 101. For the first term, 504 + x/2 + y/2, to have a factor of 101, x / 2 + y / 2 would have to be 1, for which the (x, y) possibilities are (0, 2), (2, 0) or (1, 1), but none of these results in another factor of 101 appearing in the other terms. No single-digit values for x and y can make the second term a multiple of 101. So it follows that 301 − x / 2 + y / 2 and 204 + x / 2 − y / 2 must both be divisible by 101, ie, 303 and 202 respectively. This requires y / 2 − x / 2 = 2, so y = x + 4. Since x and y are single digits, the only possibilities are these:
 x y a b c s s − a s − b s − c 0 4 505 203 304 506 1 303 202 1 5 505 204 305 507 2 303 202 2 6 505 205 306 508 3 303 202 3 7 505 206 307 509 4 303 202 4 8 505 207 308 510 5 303 202 5 9 505 208 309 511 6 303 202
As before, there must be an even number of each prime factor in s(s − a)(s − b)(s − c). Looking at the values of s, it can readily be seen that x = 1 is the only case which satisfies this condition: (3 * 13 * 13)(2)(3 * 101)(2 * 101). We therefore have x = 1 and y = 5, producing z2 = 71 and y3 = 305, thus completing the clues.

All cells in the grid can thus be completed except for the shaded areas. All complete rows and columns at this stage consist of triangular numbers, ie, they have the form n(n + 1) / 2. Solvers needed to fill the blank cells to ensure all rows and columns had this property.

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