WE CAN EXPRESS 2009 AS THE SUM OF FOUR DIFFERENT NUMBERS EACH OF...

11. We can express 2009 as the sum of four different numbers each of which consists

of at least two digits and all the digits are identical, 2009=1111+777+88+33.

What is the minimum number of addends needed to express 9002 in the same

manner?

【Solution】

Let the sum of the four-digit numbers be 1111k, the sum of the three-digit numbers be

111m and the sum of the two-digit numbers be 11n. Each of k, m and n is at most

1+2+3+…+9=45. We have 9002=11(101k+10m+n)+m. Dividing by 11, we have

k m n m −

818 101 10 4

= + + + .

11

m −

Now 4

must be some non-negative integer q, so that m=11q+4.

Since m ≤ 45, we have q ≤ 3.

Now 818=101k+10(11q+4)+n+q, so that 778=101(k+q)+10q+ n.

Since q ≤ 3, 10q+n ≤ 75 so that we must have k+q=7.

Now 10q+n=71, so that q =3, k=4, n=41 and m=37.

We have n=41=1+2+3+…+9−4. We can either take out 4 or take out 1 and 3.

We choose the latter because we want to minimize the number of terms in the sum.

Similarly, m=37=1+2+3+…+9−8, and the best result is obtained by taking away 1, 2

and 5. It follows that 9002=4444+333+444+666+777+888+999+22+44+55+66+77+

88+99, for a total of 14 terms.

ANS: 14