If you find the sum of any two-digit number and the number formed by reversing its digits, the resulting value is always divisible by what two positive whole numbers?
Show work/examples if possible
2007-03-21 16:14:59 · 3 answers · asked by Dr. Richards 2 in Science & Mathematics ➔ Mathematics
1 and 11
2007-03-21 16:17:49 · answer #1 · answered by just curious (A.A.A.A.) 5 · 0⤊ 0⤋
Let x: be the digit in the ten's place and
let y: be the digit in the one's place.
then the two digit number is:
10x + y
The number made by reversing the digits is then:
10y + x
Therefore the sum of the number and its reverse is:
(10x + y) + (10y + x)
combining similar terms:
(10x + x) + (10y + y)
giving:
11x + 11y
factoring this gives:
11(x + y)
By analyzing this answer, we can see that its factors are 11 and (x + y).
Therefore, we can say that it is always divisible by 11 and the sum of the digits.
For example:
Given number 27. Reversing the digits gives 72.
72+27 = 99
99 is divisible by 11,( 99/11= 9)
Adding the digits of the number, that is, 2+7 gives 9.
We can see that 99 is also divisible by 9.
To check if this is not just a coincidence, let's try another one.
Take the number 49. It's reverse is 94. The sum of 94 and 49 is143. 143 is divisible by 11, (143/11 is13). 4+9 =13. 143 is also divisible by 13.
Therefore, we can conclude that the sum of any two digit number and the number made by reversing it's digits is always divisible by 11 and the sum of the digits.
2007-03-21 16:36:05 · answer #2 · answered by activista 2 · 0⤊ 0⤋
98+89 = 187 = 11*17
16 + 61 = 77 = 11*7
35 + 53 = 88 + 11*2*2*2
51 + 15 = 66 = 11*3*2
notice anything? all are divisible by 11! and any number is divisible by 1, so the answer is 1 and 11.
2007-03-21 16:23:24 · answer #3 · answered by mitch w 2 · 1⤊ 0⤋