26. Find the smallest and the largest three digit numbers which when divided by 22,33 and 55 leave a remainder of 5 in each case. 01) 340, 980 O2) 335, 995 O 3) 330,990 0 4) 325, 985
Answers
Answered by
0
22-5=17
33-5=28
55-5=50
Prime factorization of 17=1×17
Prime factorization of 28=1×2×2×7
Prime factorization of 50=1×2×5×5
HCF is 1
LCM is 2×2×5×5×7×17=11900
I think I have answered correctly.
Answered by
1
Answer:
the answer
Step-by-step explanation:
Prime factorisation of
22=
2×
11
Prime factorisation of
33=
3×
11
Prime factorisation of
55=
5×
11
So, LCM
22,33,55=
2×
3×
5×
11=
330
As
330 is the smallest 3− digit number divisible by
22,33,55
, the number
330+
5=
335
will give a remainder
5
when divided by these numbers.
The highest
3− digit number is 999 .999 when divived by 330 gives a remainder 9
, so
999−
9=
990
is the largest
3−
digit number divisible by
22,33,55
, the number
990+
5=
995
will give a remainder
5
when divided by these numbers.
Similar questions