Solutions to Jumira's Brain-Challengers

1. A and B are numbers such that AB = 1 000 000. Find A and B, if neither of them contain zeros.

Solution: (given by O Magwaza, L Hoba, I Warirai, and T Nhakai of Hippo Valley High School, Chiredzi)

AB = 1 000 000     =    106     =    (2×5)6 = (26)×(56)     =    64×15625

The values of A and B are 64 and 15625. Both of them contain no zeros.

(This problem was also solved by K Kundanayi of Muchekayaora Secondary School, Gutu, and by O Sibanda and J Shumba of Zvishavane)

2. Find a four digit number which is exactly four times greater when its digits are reversed.

Solution: (given by T Mutingwende of Mount Pleasant, Harare)
Suppose the digits are A, B, C and D in that order. Then

1000A + 100B + 10C + D = 4000D + 400C + 40B + 4A Û 322A + 20B -130C - 1333D = 0.

From this, we see that D (the last digit) is even and (since the heavyweight terms with A and D have to nearly balance each other, and 4×322 » 1333) A is likely to be 4D. This leads us to A and D being 8 and 2 respectively. Therefore

332(8) + 20(B) -130C -1333(2) = 0      or       2B - 13C = 1.

Now we try digits from 0 to 9 for B and find one which gives the possible value of C. The only possible value of B is 7 which gives the value of C as 1. Therefore the number is 8712 which gives 2178 when divided by 4.

Another Solution: (given by T Masiwa of Rufaro Secondary School, Neshuro)
Let the number be abcd such that 0 £ a, b, c, d £ 9. But a ¹ 0 and d ¹ 0. From the fact that 4(abcd) = dcba, we obtain the following equations

4d = 10n + a          (1) 4a + x = d           (2) 4c+n = 10m + b           (3) 4b + m = 10x + c           (4)

Now, n, m, x are integers such that 0 £ n, m, x £ 3 since 9×4 = 36. From equations (1) and (2), we get 4(4a + x) = 10n + a, from which we obtain that

a = 10n - 4x 15 .

But 10n - 4x is divisible by 15 only when x = 0 since 0 £ x £ 3. Hence, from equations (3) and (4), we have

4(4b+m) + n = 10 m + b       or       b = 6m-n 15 .

Since 0 £ m, n £ 3,  6m-n is divisible by 15 only for m = n = 3. Thus we get:  a = 2,     b = 1,     c = 7,    d = 8. Hence the numbers are 2178 and 8712.

Another Solution (given by O Sibanda and J Shumba of Zvishavane)
Let the four digit number be abcd. Since 4(abcd) = dcba, the right hand side must be even and this is only possible if a is even. The only possible solution is a = 2 as the product of abcd and 4 is also a four digit number.

Now, 4×(abcd) = dcba Þ 1333a -332d = 20c -130b,  as in previous solution (with letters in reverse order). With a = 2, we see that this equation becomes 2666 - 322d = 20c - 130d. The left hand side of this equation must be divisible by 10 as the right hand side is divisible by 10. This is possible only if the last digit of 322d is 6. From the first step in the above array, we see that d ³ 4a i.e., d ³ 8. Therefore d = 8. Substituting this value of d into the equation 2666 - 332d = 20c - 130b we obtain 1 = 2c - 13d. The only possible solution to this equation is b = 1 and c = 7. Therefore the four digit number is 2178.

(This was also solved by O Magwaza, L Hoba, I Warirai and T Nhakai of Hippo Valley High School, Chiredzi)

3. Find the digits MANGO if (M + A + N + G + O)³ = MANGO.

Solution: O Magwaza, L Hoba, I Warirai, T Nhakai of Hippo Valley Secondary High School, Chiredzi used the sure but long searching method:

Therefore MANGO is 17576 and 19683.

(K Kundanayi of Muchekayaora Secondary
School, Gutu, and O Sibanda and J Shumba of Zvishavane, provided some analysis which shows just where to search:)
Ö[3]99  999 = 46,4¼ and Ö[3]10  000 = 21,5¼. Therefore a whole number x whose cube gives us a 5-digit number is such that 22 £ x £ 46, from which we easily find 26³ = (1+7+5+7+6)³ = 17576. Therefore MANGO = 17576 is a solution.
Sibanda and Shumba go on to observe that only the numbers: 24, 27, 29, 32, 35, 38 and 41 give cubes with five different digits as shown below

24³ = 13824        27³ = 19683        29³ = 24389        32³ = 32768        35³ = 42875        38³ = 54872        41³ = 68921

Only 27 is equal to the sum of digits of its cube. Therefore the only solution with distinct digits is MANGO = (1+9+6+8+3)³ = 27³ = 19683.

Masiwa, in evaluating the cubes of integers between 22 and 46, observed that the sum of 26 occurs every third integer from 26, that is, for 26³, 29³, 32³, ¼, 44³.

4. Show (without using a calculator or tables) that \dscos2001^° > -[1888/1999].

Solution: (given by J Shumba and O Sibanda of Zvishavane)
2001^° has the same cosine as q where 0^° < q < 180^° and q satisfies the following:

360° n - q = 2001°,    n  is   a   natural   number.

Thus

q = 360° n - 2001° = 2160° - 2001° = 159°     for   n = 6.

Therefore we need to show that

cos159° > - 1888 1999    i.e.  -cos21° > - 1888 1999    i.e.  cos21° < 1888 1999 .

We can show this by constructing a right-angled triangle ABC with BC = 100 mm, angle B = 90^° and angle C = 21^°.. An accurate construction of this triangle shows that

cos21° » 100 107 < 1888 1999 .

Therefore cos21^° > -1888/1999.

An Exact Solution not involving `accurate construction': ( given by the setter Oswald Jumira)
2001 = 5×360 + 201 and so cos2000^° = cos201^° = - cos21^°. By the concavity of the cosine function on [0^°, 30^°] (this means that a line segment drawn between any two points on the graph will lie wholly underneath the curve), we have

cos21° = cos æ ç è 9 30 ×0° + 21 30 ×30° ö ÷ ø £ 9 30 cos0° + 21 30 cos30° = 9 30 + 21 30 × Ö3 2 = 18 + 21Ö3 60 < 1888 1999

Thus cos2001^° > -\ds[1888/1999].

5. A six digit number x1996y is divisible by 99. Find x and y.

Solution: (given by T Masiwa of Rufaro Secondary School, Neshuro)
Since x1996y is divisible by 99, then x1996y is divisible by 9. Thus x+y+25 is divisible by 9. Hence

x+y = 2      OR     x+y = 11,

as x,y Î {0,1,2, ¼, 9}. Moreover, x1996y is also divisible by 11. Thus x - 1 + 9 - 9 + 6 -y is divisible by 11 by the `Charles Dodgson' test for divisibility by 11. Hence x-y+5 is divisible by 11. This is only possible when x-y+5 is 0 or 11, because x,y Î {0,1,2, ¼, 9}. That is:

x-y = -5     OR      x-y = 6 .

Of the four possible pairs of simultaneous equations for x, y, three pairs are inconsistent. Using the only consistent pair x+y = 11,  x-y = -5, we get x = 3   and   y = 8. Thus x1996y = 319968 = 99×3232.

Another Solution: (given by O Sibanda and J Shumba of Zvishavane)
Since x1996y is divisible by 99, it is divisible 9 and by 11. From the fact that it is divisible by 9, we know that the sum of the digits is a multiple of 9. The only possible sum of the digits is 36. From this, we obtain the equation

x + y = 11.

Now, using divisibility by 11, we have (10x +1) + 99 + (60 +y) = multiple   of  11. This reduces to

10x + y = 11n -160,

where n is a natural number. Eliminating y from the two equations, we obtain 9x = 11n - 171. Taking into account that x is a natural number such that 0 £ x £ 9, we get x = 3. From this we obtain that y = 8. Therefore the number is 319968.

(Correct solutions by mainly trial and error were given by O Magwaza, L Hoba, I Warirai and T Nhakai of Hippo Valley High School, Chiredzi and by K Kundanayi of Muchekayaora Secondary, Gutu)

6. In triangle ABC, angle A is twice angle B, angle C is obtuse and the lengths of the three sides are integral units. Determine, with proof, the minimum possible perimeter.

Solution: (given by T Masiwa of Rufaro Secondary School, Neshuro)
Refer to the diagram above. Let x, a, b be natural numbers. We wish to find the relationship between x, a, b and a. Using the sine rule, we have

x sina = x+a sin2a = x+b sin3a .

Hence

xsin2a sina = x+a      and      xsin3a sina = x+b.

From the first of these, we obtain    x(2cosa- 1) = a,
and from the second we get

2x(2cos2a- 1) = b       or       4xcos2a- 2x = b.

If b is a natural number, then 4xcos²a is a natural number greater than 2x.

Let cosa = t/m where a < 30^°, t < m and t and m are relatively prime natural numbers. Thus 0 < cosa < 1. It follows that 4x is divisible by by m². To minimise the perimeter, x has to be small as possible. 4x ³ m². So if we take 4x = m², m must be even. Hence we need the smallest even number m which satisfies the condition:

\dscos-1[(m-1)/m] < 30° ,    since \ds cos-1 [(m-1)/m] < cos-1[(m-2)/m] < ¼.

Consider the table below.

We see that m = 8 and so a = 12,  b = 17,  x = 16 and the perimeter p = 16+28+33 = 77.
Alternatively, observe that cosine is a decreasing function on [0,90^o], and

1 2 < 3 4 < 5 6 < Ö3 2 = cos30o < 7 8 .