Play with Two Numbers
Find two numbers a and b having these conditions: their sum is 1244; the numbers would be equal if the digit 3 were added in the end of the first number, and the digit 2 erased from the end of the second number.
Write your answer in the form of ⌊ a b ⌋ .
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3 solutions
The problem can be solved without using equations. First of all, we have to notice that the first number is 2-digit number and the second is 4-digit number. Then we come to the conclusion that the second number must have the last digit of 2 and the third digit of 3. The knowledge about the sum of two numbers brings us to the solution: a = 12 and b= 1232. Some of my 5th - grader kids successfully solved this problem by such a way, but creating the second equation is a good and even easier way to find the numbers.
As the answer must be given as floor (b/a), b must be larger than a.
As (the number of digits of a) + 1 equals (the number of digits of b) - 1, a must have two digits less than b . Combined with the fact that they make 1244 together, a must consist of 2 digits and b must consist of 4 digits.
Call the digits of b ABCD and call the digits of a EF.
D = 2 (as given) and A must be 1 (else they can't add up to 1244), so a = 1BC2.
1BC = EF3 (as given), so C = 3 and E = 1, so a = 1B32 and b = 1F.
1B32 + 1F = 1244, so F must be 2 and B must be 2, so a = 1234 and b = 12
Given that a + b = 1 2 4 4 ⟹ b = 1 2 4 4 − a . . . ( 1 ) .
a added a 3 in the end ⟹ 1 0 a + 3 . Digit 2 is erased from the end of b ⟹ 1 0 b − 2 . Therefore
1 0 a + 3 ⟹ 1 0 0 a + 3 0 + 2 1 0 0 a + 3 2 1 0 1 a ⟹ a b = 1 0 b − 2 = b = 1 2 4 4 − a = 1 2 1 2 = 1 2 = 1 2 4 4 − 1 2 = 1 2 3 2 Since ( 1 ) : b = 1 2 4 4 − a .
Therefore, ⌊ a b ⌋ = 1 0 2 .