I AM A TWO DIGIT NUMBER LESS THAN 50. IF YOU PUT ME IN GROUPS OF 7, THEN THERE ARE 2 LEFT OVER. THE SUM OF MY DIGITS IS 8. WHO AM I ?
To find the two-digit number that satisfies the given conditions, we can use a systematic approach and try different numbers until we find one that matches all the criteria.
Let's start by listing all the two-digit numbers less than 50 and examining their properties.
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49
To determine if a number can be divided into groups of 7 with 2 left over, we need to check if the number is congruent to 2 modulo 7 (i.e., leaves a remainder of 2 when divided by 7). Let's filter out numbers that do not satisfy this condition:
11, 18, 25, 32, 39, 46
Next, we need to find the numbers for which the sum of the digits is 8. Let's examine each of the remaining numbers:
11 is not a valid option as the sum of its digits is 1 + 1 = 2.
18 is a valid option as the sum of its digits is 1 + 8 = 9.
25 is not a valid option as the sum of its digits is 2 + 5 = 7.
32 is not a valid option as the sum of its digits is 3 + 2 = 5.
39 is not a valid option as the sum of its digits is 3 + 9 = 12.
46 is not a valid option as the sum of its digits is 4 + 6 = 10.
From this analysis, we can conclude that the only two-digit number that satisfies all the conditions (less than 50, leaves a remainder of 2 when divided by 7, and has a sum of digits equal to 8) is 18.
Therefore, the answer to the riddle is 18.