2006-07-18 20:00:43 · 15 answers · asked by Babli 2 in Science & Mathematics ➔ Mathematics
To answer Hi Y'all's question, 27 is the only positive integer with this property. To prove this, let a_k a_(k-1) ... a_1 a_0 be the digits of the integer written in base 10 (so it is a k+1 digit number, and I am assuming that a_k is not 0). Then 3 times the sum of the digits is
3a_k + 3a_(k-1) ... 3a_1 + 3a_0, while the number itself is given by a_k*10^k + a_(k-1)*10^(k-1) + ... +a_1*10^1 + a_0. If these expressions are equal, we have
3a_k + ... + 3a_0 = a_k*10^k + ... + a_0, so that
(10^k-3)a_k = -(10^(k-1) - 3)a_(k-1) - ... - (10-3)a_1 + 2a_0
Now clearly each of the terms (10^i-3)a_i is nonnegative for i running between 1 and k. Thus, the sum on the right is less than or equal to 2a_0, so that (10^k-3)a_k <= 2a_0 <= 18 since a_0 is a digit, so is no greater than 9. On the other hand, we assumed that a_k was not 0, and it is a digit as well, so it is between 1 and 9. Thus, (10^k-3)a_k >= (10^k-3), so that (10^k-3) <= 18. The only way that this can happen is if k=1 or k=0, so that the number has 1 or 2 digits.
If k=1, then the equality giving the property of the number we are interested in is a_1*10 + a_0 = 3a_1 + 3a_0, so that 7a_1=2a_0. Since both sides are integers, unique factorization shows that 7 divides a_0, and since a_0 is a single digit, it must be 7. Then the only possibility for a_1 is that it is 2, so the number is 27.
If k=0, then the equality reads a_0 = 3a_0, which is impossible unless a_0=0, but we assumed the leading digit of our number was nonzero.
Thus, the only positive number with this property is 27. Of course, the respondant who answered 0 was also correct.
2006-07-18 20:24:49 · answer #1 · answered by mathbear77 2 · 1⤊ 0⤋
27
2006-07-18 23:42:57 · answer #2 · answered by budweiser 2 · 0⤊ 0⤋
27
2006-07-18 20:50:07 · answer #3 · answered by Rohit C 1 · 0⤊ 0⤋
27
2006-07-18 20:04:58 · answer #4 · answered by jazzmen4u28 3 · 0⤊ 0⤋
27
2006-07-18 20:03:27 · answer #5 · answered by just_beju 2 · 0⤊ 0⤋
27
2006-07-18 20:02:09 · answer #6 · answered by Steve W 3 · 0⤊ 0⤋
6 or 2
2006-07-18 20:07:54 · answer #7 · answered by sloarbag 1 · 0⤊ 0⤋
0 =3*0
27 =3*(2+7)
numbers with 3 (or more) digits won't work, because the biggest sum of digits would be 27=9+9+9. and 3*27 < 100.
negative numbers won't work either, because the sum of digits is always positive.
fractional numbers like 123.45 won't work, because the sum of digits is a natural number => three times that is natural as well.
2006-07-18 20:15:45 · answer #8 · answered by yushoor 1 · 0⤊ 0⤋
27 it is!
2006-07-18 22:07:26 · answer #9 · answered by early_sol 2 · 0⤊ 0⤋
One solution would be 27, but is it the only one? I wonder. Unfortunately it would take too much time to find out.
2006-07-18 20:04:38 · answer #10 · answered by Hi y´all ! 6 · 0⤊ 0⤋