IPMAT 2022 - Quantitative Aptitude Question 27
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IPMAT 2022 - Quantitative Aptitude
Four digit numbers divisible by both 2 & 3 would be numbers divisible by 6
Total numbers divisible by 6 & > 1000 = 1002, 1008, 1014 ….. 9996
an = a + (n – 1)d
9996 = 1002 + (n – 1) 6
8994/6 = (n – 1)
n = 1500
Number divisible 2, 3 & 5 = Numbers divisible by 30
Total numbers divisible by 30 = 1020, 1050 ….. 9990
an = a1 + (n – 1)d1 a1 = 1020 d1 = 30
9990 = 1020 + (n – 1) 30
8970/30 = n-1
n = 300
So number of 4 digit numbers greater than 1000
divisible by both 2 & 3 but not 5 = 1500 – 300 = 1200