Problem 170 — Spare the rod and you can end up spoiling the whole show, as this month’s problem by E.A. Engebretson of St. Paul, Minn., demonstrates.
It was a quarter to 8:00 a.m. on the grand opening day of the new Mammoth Mart shopping complex. Store manager Eustace Snipp was adjusting his bow tie preparatory to having his picture taken cutting the red ribbon in front of the store.
Suddenly, a resounding crash echoed through the aisles of the new Mammoth Mart shopping complex. The letter “o” in the glass sign proclaiming the store name had fallen, and a pie-shaped wedge with a circular arc of 45 deg and 10 in. had broken off the top of the letter.
With a volley of ‘Oh, dear’s and ‘Oh my’s Snipp raced to the scene. The workmen were just raising the fallen letter when Snipp got an idea.
“We’ll just bend this rod over the top and fit the broken glass back in. From a distance no one will be able to tell.”
The rod is a thin strip of steel, 1/32-in. and long enough so it can be clamped at each end and span the 10-in. circular gap. Its modulus of elasticity is 30 x 106. The rod has to be used later, and so it has to be kept in elastic bending. Thus, the bending stress is limited to 49,000 psi. What is the maximum angle of circular arc that can be maintained by the rod? Will Snipp’s grand opening go off with a bang? Assume the rod is in pure bending, and the weight of the refitted broken glass is negligible.
Technical consultant, Jack Couillard, Menasha, Wis.
Solution to last month’s problem 169 — You’re right on the money if you answered $9,567 for SEND, $1,085 for MORE, and $10,652 for MONEY. Here’s the final balance:
We know that M = 1, because 0 is excluded and MO,NEY < 19,998 (9,999 x 2)
Then, since S 1 1 1 a possible carryover of 1 from column E 1 0 5N has to equal 10 (11 is not possible because 1 is already used, and 12 or more is not reachable.)
O = 0. S has to equal 8 or 9. If S = 8, then we would need E + O + a carry-over to be >12, which, again, is not possible.
Therefore, S = 9, and:
The largest number that can be carried over from any column is 1. Therefore we can deduce that E + 1 = N. We are then left with the following unknowns:
From this, we can set up the following algebraic equation:
100E + 10(N + R) + D + E = 100N + 10E + Y
Substitute E + 1 for N:
100E + 10(E + 1 + R) + D + E = 100(E + 1) + 10E + Y
10R + D + E = 90 + Y
Therefore, R = 8. 9 is already used, and so it is not possible for R < 7 to satisfy 10R + D 1 E 590 1 Y, since 70 + D + E > 90 implies D + E > 20, which is not possible.
Again, we can set up an algebraic equation. Plugging R = 8 into our last equation:
80 + D + E = 90 + Y
D + E = 10 + Y
Now since 8 and 9 are already used, the highest remaining numbers are 6 and 7.
D + E < 13
Y = 2 or 3
If Y = 3, then D + E =13. Therefore, with the numbers 3, 4, 5, 6, and 7 remaining, D = 6 or 7; and E = 6 or 7. But E + 1 = N, so E ≠ 6 or 7. Y, then, has to be 2.
D + E = 12. By elimination: E = 5; D = 7; and N = 6.
The solution is then:
The customers were certainly sending more money!