1,257
1,275
1,527
........
........
+7,521
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum R of the 24 integers ? Choices:	A) 26,996	B) 44,406	C) 60,444	D) 66,660	E) 99,990
This is the way i solved it :   Total no. of different combos = 24, total digits 4 therefore each digit must repeat  24/4= 
<gadget id="calculator"> 24/4</gadget>
<output>6</output>
6 times in each row .. So the right most row would add up to 1x6 + 2x6 + 5x6 + 7x6= 6+12+30+42= 
<gadget id="calculator">6+12+30+42</gadget>
<output>90</output>
90 .. Each row would add up to 90, so 90 in the first means we have 9 that carries over and we get 0 , the second time its 90+9 and 9 stays and one 9 goes to the row to the left, so the last two digits of the SUM R should be 90 (E) .. We could go on and solve the exact number but since only one answer choice has the last digits as 90 we needn't go any further..
<result>E</result>
1,257
1,275
1,527
........
........
+7,521
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum R of the 24 integers ?