Week 20

Example 1. Given that 1000002000001 has a prime factor greater than 9000. Find it.

Example 2. Let A = {(a, b, c) : a, b, c are prime numbers satisfying a < b < c and a + b + c = 30}. Find the number of elements in A.

Example 3. Let ABCDEFGHIJ be a 10-digit number, where all the digits are distinct. Further, A > B > C, A + B + C = 9, D > E > F > G are consecutive odd numbers and H > I > J are consecutive even numbers. Find the value of A.

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. Seeing the given number 1000002000001, it’s difficult to proceed further. But a closer look might help us. We can write

1000002000001

= 1000000000000 + 2000000 + 1

= 1012 + 2 × 106 + 1

= (106)2 + 2 × 106 + 1

= (106 + 1)2

= [(102)3 + 13]2

= [(102 + 1){(102)2 − 102 + 1}]2

= (102 + 1)2[(102)2 − 102 + 1]2

= (100 + 1)2[10000 − 100 + 1]2

= 1012 × 99012.

Therefore, the prime sought is 9901.

Solution 2. In essence the question asks in how many ways 30 can be expressed as a sum of three distinct primes.

Now, 30 is an even number and we know that the sum of three even numbers is odd. Not all three of these primes can be odd. Following table shall help us.

abca + b + c
EvenEvenOddOdd
EvenEvenEvenEven
EvenOddEvenOdd
EvenOddOddEven
OddEvenEvenOdd
OddEvenOddEven
OddOddEvenEven
OddOddOddOdd

Since 2 is the only even prime and a < b < c, all of a, b and c cannot be even.

Now consider a = 2 and hence b + c = 28. Since b < c, the only possible values b can take are 3, 5, 7, 11 and 13. From these assumed values of b, c = 28 – b is also a prime only when b = 5 and 11.

So there are only two members in A and they are (2, 5, 23) and (2, 11, 17).

Solution 3.  The only odd digits are 1, 3, 5, 7 and 9. Since A + B + C = 9, an odd number, either exactly one or all three of A, B, C are odd.

Let A, B, C be all odd digits and given that D, E, F, G are consecutive odd numbers, this is not possible. So, only one of A, B, C is odd. Rest two are even.

The only way we can have four consecutive odd digits in a descending order is 9 > 7 > 5 > 3 and 7 > 5 > 3 > 1. If we consider the latter possibility, then 9 must be any one of A, B, C. Then the remaining two of these three will add to 0, since A + B + C = 9.

Hence both will be 0, which is not allowed as all the letters represent distinct digits. So we conclude that D = 9, E = 7, F = 5, G = 3 and 1 is any one of A, B, C. So the remaining two add to 8.

Next we focus on the last condition, that H, I, J are consecutive even numbers with H > I > J. This time there are three possibilities:

(i) 8 > 6 > 4                (ii) 6 > 4 > 2   and      (iii) 4 > 2 > 0.

Recall that, any two of A, B and C are even. In each case, the missing two even digits are in A, B, C and add to 8. The even numbers left in (i) are 2, 0 and 2 + 0 = 2. Similarly, the even numbers left in (iii) are 8, 6 and 8 + 6 = 14. This rules out (i) and (iii). So we are left with (ii).

Now the missing two even digits 8 and 0 are in A, B, C. Hence A, B, C are 1, 0 and 8. As they are given to be in a descending order, we have A = 8.

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