+0

# number theory

0
72
1

How many different positive integers divisible by 4 can be formed using at least one of the digits 1, 2, 3, 4, and 6 exactly once and no other digits? For example, 12 counts, but 512 does not.

Jan 7, 2021

#1
0

4 by itself = 1 such number.

(12, 16, 24, 32, 36, 64)>>Total = 6 [These are 2-digit numbers that are divisible by 4]

(124, 132, 136, 164, 216, 236, 264, 312, 316, 324, 364, 412, 416, 432, 436, 612, 624, 632) >>Total = 18 [These are 3-digit numbers divisible by 4]

(1236, 1264, 1324, 1364, 1432, 1436, 1624, 1632, 2136, 2164, 2316, 2364, 2416, 2436, 3124, 3164, 3216, 3264, 3412, 3416, 3612, 3624, 4132, 4136, 4216, 4236, 4312, 4316, 4612, 4632, 6124, 6132, 6312, 6324, 6412, 6432) >>Total = 36 [These are 4-digit numbers divisible by 4]

(12364, 12436, 13264, 13624, 14236, 14632, 16324, 16432, 21364, 21436, 23164, 23416, 24136, 24316, 31264, 31624, 32164, 32416, 34216, 34612, 36124, 36412, 41236, 41632, 42136, 42316, 43216, 43612, 46132, 46312, 61324, 61432, 63124, 63412, 64132, 64312) >>Total = 36 [These are 5-digit numbers divisible by 4]

So, the total =1 + 6 + 18 + 36 + 36 =97 such permutations

Jan 7, 2021