The total number of numbers lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if repetition of digits is not allowed, and numbers are divisible by either 3 or 5, is ...........
Answer: 32
Solutions · 1
Official solution
Digits: {1,2,3,4,5}, forming 3-digit numbers without repetition.
Divisible by 3 (digit sum divisible by 3): valid digit triples: {3,4,5}(sum12), {2,3,4}(sum9), {1,3,5}(sum9), {1,2,3}(sum6) --- 4 triples, each giving 3!=6 arrangements: 24.
Divisible by 5 (ending in 5): remaining 2 digits from {1,2,3,4} arranged: 4×3=12.
Divisible by 15 (both): numbers ending in 5 with digit sum divisible by 3: 145,415,345,435 --- 4 numbers.
\[\text{Total} = 24+12-4 = 32\]