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math problem

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How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

Sep 8, 2020

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1025 , 1035 , 1045 , 1205 , 1230 , 1235 , 1240 , 1245 , 1250 , 1305 , 1320 , 1325 , 1340 , 1345 , 1350 , 1405 , 1420 , 1425 , 1430 , 1435 , 1450 , 1520 , 1530 , 1540 , 2015 , 2035 , 2045 , 2105 , 2130 , 2135 , 2140 , 2145 , 2150 , 2305 , 2310 , 2315 , 2340 , 2345 , 2350 , 2405 , 2410 , 2415 , 2430 , 2435 , 2450 , 2510 , 2530 , 2540 , 3015 , 3025 , 3045 , 3105 , 3120 , 3125 , 3140 , 3145 , 3150 , 3205 , 3210 , 3215 , 3240 , 3245 , 3250 , 3405 , 3410 , 3415 , 3420 , 3425 , 3450 , 3510 , 3520 , 3540 , 4015 , 4025 , 4035 , 4105 , 4120 , 4125 , 4130 , 4135 , 4150 , 4205 , 4210 , 4215 , 4230 , 4235 , 4250 , 4305 , 4310 , 4315 , 4320 , 4325 , 4350 , 4510 , 4520 , 4530 , 5120 , 5130 , 5140 , 5210 , 5230 , 5240 , 5310 , 5320 , 5340 , 5410 , 5420 , 5430 , Total =  108 such numbers.

Beginning with each of the 4 numbers, 1, 2, 3, 4, each number begins with 4! =24 permutations. 24 x 4 = 96 permutations. Beginning with 5 itself, ordinarily you would also have 4! or 24 permutations, but because you cannot have 2 "fives" in the same permutation, then we can only have: 24/ 2 = 12 permutations beginning with 5.

So: 96 + 12 = 108 such numbers.

Sep 8, 2020