Week 27

Example 1. What is the smaller angle between the two hands of a clock when it shows exactly 7 : 22?

Example 2. The product of four positive integers a, b, c and d is 8!, and they satisfy the equations

ab + a + b = 524,
bc + b + c = 146, and
cd + c + d = 104.

Find the value of a − d.

Of course, you can find the solution just below, but it is highly recommended that you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. When it is 7 : 00, the largest possible angle between the hands is (7/12) × 360° = 210° (see the figure).

In each hour,

the minute hand moves 360°; and
the hour hand moves (1/12) × 360° = 30°.

Thus in 22 minutes,

the minute hand has moved (22/60) × 360° = 132°; and
the hour hand has moved (22/60) × 30° = 11°.

Therefore, the new angle between the hour and minute hands will be

210° − 132° − 11° = 89°.

Solution 2. A critical observation in the solution to this problem is that the three equations can be rewritten as

525 = 524 + 1 = ab + a + b + 1 = (a + 1)(b + 1),
147 = 146 + 1 = bc + b + c + 1 = (b + 1)(c + 1), and
105 = 104 + 1 = cd + c + d + 1 = (c + 1)(d + 1).

Now factor the constant terms in each equation to obtain facts about the products.

(a + 1)(b + 1) = 525 = 3 · 5² · 7,
(b + 1)(c + 1) = 147 = 3 · 7², and
(c + 1)(d + 1) = 105 = 3 · 5 · 7.

Since (a + 1)(b + 1)has a factor of 5² = 25, but (b + 1)(c + 1)has no factor of 5, (a + 1)must be divisible by 25. In a similar manner, (d + 1) must be divisible by 5.

Because (b + 1)(c + 1)= 3 · 7², the possibilities for (b + 1)and (c + 1) are either (b + 1)= 7 and (c + 1)= 3 · 7 or are (b + 1)= 3 · 7 and (c + 1)= 7.

However, if (b + 1)= 7, then (a + 1)= 3 · 25 = 75 and a = 74. But we are given that a · b · c · d = 8!, and 74 does not divide 8!.

So, we cannot have (b + 1)= 7. Hence we must have (b + 1)= 3 · 7 and (c + 1)= 7. This implies that (a + 1)= 25 and (d + 1)= 3 · 5 = 15. Thus a = 24, b = 20, c = 6, and