4 ms·
multiples of 11 => you need to get the digits to sum to be odd => you need to find a number with a 2:1 ratio of even:odd digits {odd + odd = even, even + even =
by zackattack 16y ago
multiples of 11 => you need to get the digits to sum to be odd => you need to find a number with a 2:1 ratio of even:odd digits {odd + odd = even, even + even = even, odd + even = odd, odd + even + even = odd [(odd + even)=odd + even = odd], etc.}.
so 209 is the first number with the appropriate 2:1 ratio (edited).
- dawgr 16y ago1+5+4=10 so that's not it. if you meant 154x11=1694 1+6+9+4=20 so that's not it either. By checking manually, it's 209. 19x11=209 2+0+9=11 The way you are doing doesn't seem correct even if what you said it's true, you aren't multiplying by 11. I have no idea what the reasoning to prove the negative would be. I'd be interested to read it, if anyone knows.
- zackattack 16y agoThe reason I began my post with "multiples of 11 =>" is because I was addressing that problem.
- kleevr 16y ago1+5+4 = 10 (even digit sum) For small values of X < 10 11X = X+X, 2X always even For slightly larger values of X, where 11X < 200 11x10 = 110, (starts out even) 11x11 = 121, (both of the first two digits increment by one, flipping even or odd ... in unison) 11x18 = 198, (only two signs will flip until we increment the 100s place, when that happens we will reverse three signs instead of two) 11x19 = 209, (now, three instead of two flip, and our sum comes up odd for the first time, 2+0+9=11)