I am a 2-digit number over 50. When you put me in groups of 7, two are
left over. The sum of my digits is 11. What number am I?
7 x + 2 = integer multiple of x
well, over 50 so list
49+2 = 51
56+2 = 58
then keep adding 7
65
but 6+5 = 11 score !
7*9 + 2
70
Well, well, well. If you're a 2-digit number over 50, then you must be a sweet little number with some maturity. Let's put on our thinking caps!
Now, suppose we start brainstorming. To put you in groups of 7, two are left over, which is a bit odd. But hey, I'm a clown, I love odd things!
Now, think about the sum of your digits being 11. That's like adding a juggling ball with a bowling ball! It sounds like quite the balancing act.
So, after some clownish calculations, the number that fits the bill is... drum roll, please... 61!
You're a charming little number indeed!
To find the number, we can start by figuring out the possibilities for the sum of the digits, since we know that it is 11.
To make a 2-digit number with a sum of 11, the only possible combinations are 29 and 38.
Next, we need to find the 2-digit number over 50 that, when divided by 7, leaves a remainder of 2.
To do this, we can start with the smaller number, 29, and add multiples of 7 until we find a number that fits the criteria.
29 + 7 = 36 (not greater than 50)
29 + 2(7) = 43 (not greater than 50)
29 + 3(7) = 50 (not greater than 50)
Since none of these numbers are greater than 50, we move on to the next possibility, 38.
38 + 7 = 45 (not greater than 50)
38 + 2(7) = 52 (greater than 50)
The number that fits all the criteria is 38. Therefore, the answer is 38.