2026/048) For $1\le n \le2016$, how many integers n satisfying the condition: the remainder divided by 20 is smaller than the one divided by 16

Because LCM of 20 and 16 is 80. we need to look upto 80 an then the pattern repeats

Now remainder shall keep increasing till we get next multiple . then the remainder resets

Because we need to compare the remainder the number is of the form 20a + b and 16c + d ( 0 <=b <=19) , (o <=d <=16)

The remainder divided by 20 is lower than remainder divided by 16 if multiple of 16 is lower than multiple of 20. that is $16c < 20a < n < 16(c+1) $, That is the number lies between multiple of 20 and multiple of 16. The numbers that fit the condition are from 20a to 16(c+1)

Let us write multiples of 16 : $16,32,48,64,80$

multiple of 20 : -$20,40,60,80$

So the ranges are $0, 16(16*1),20(20*1),32(16*2),40(20*2),48(16*3),60(20*3),64(16*4),80$

the ranges where the remainder divided by 20 is lower in the rage $[20*1,16*2), [20*2,16 * 3]$ and $[20*3,16 * 4)$ in each range lower number is inclusive and higher is not giving $16*2 - 20 *1 + 16*3 - 20*2 + 16*4 - 20 *3$ or $24$ numbers.

In each lock of 80 numbers there are 24 numbers

upto 2000 there are 25 blocks so 25 * 24 600 numbers

From 2001 to 16 there are 16 number having same remainder as divided by 16 as divided 20

So no extra

So Ans 600