The smallest positive integer that appears in each of the arithmetic progressions 5, 16, 27, 38, 49,
...; 7, 20, 33, 46, 59, ....; and 8, 22, 36, 50, 64, .... Is
Answers
the smallest positive integer that appears in each of the AP is 1996
Step-by-step explanation:
number would be
11a + 5 = 13b + 7 = 14c + 8
11, 13 & 14 are co prime
and LCM =2002
so such numbers will repeat every 2002
one method to find using hit and trial
let's take 2 at a time
11a + 5 = 13b +7
11a = 13b + 2
b = 10 a = 12
b = 21 a = 25
and so on
13b + 7 = 14c + 8
13b = 14c + 8
c= 5. b = 6
c = 18. b = 20
11a+5 = 14c + 8
11a = 14c + 3
c = 10 a = 13
c = 21 a = 27
if we notice a sequence
13n -1. and 14n - 1
13×14 -1 = 181
so common first a =181 then repeat after 182
for this a = 181
b = 153. c = 142
now check
11a + 5 = 13b + 7 = 14c + 8
1996 = 1996 = 1996
1996 will be smallest integer satisfying this condition
another method
if we see series
5,16,27,38,49,....; 7,20,33,46,59...; and 8,22,36,50,64....
we see difference between numbers and difference between common difference is same
hence if we see one number back side. it will satisfy
5-11 = -6
7-13=-6
8-14=-6
LCM of 11,13& 14 = 2002
now add 2002 in -6 to get 1996
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