Question

672 = 4489
6672 = 444889
= 44448889
666672 =
(111) As the ones place of 157n
square of the number will
ones place.
Check: 7 x 7 = 49.
1. Observing the above, we get
(i) 66672 = 44448889
(ii) 666672 = 4444488889
pe 2
Without actual addition, find the sum of the
following
Note
The square of a number having:
1 or 9 at the ones place ends in 1.
2 or 8 at the ones place ends in 4.
3 or 7 at the ones place ends in 9.
4 or 6 at the ones place ends in 6.
5 at the ones place ends in 5.
Factice Questions 6B
pe 1
Without actual multiplication, find the squares of:
(i) 1111111
(ii) 215
(iii) 95
(iv) 666667​

Answer 1

Answer:

is this a question of English??

Answer 2

Heya branliy user ! your answer

Prove that in a sequence of numbers 49,4489,444889,44448889…

elementary-number-theory

Prove that in a sequence of numbers 49,4489,444889,44448889… in which every number is made by 48 in the middle of previous as indicated, each number is the square of an integer.

MathGod

(610k−19+1)2=36102k−2⋅10k+181+1210k−19+1=410k−19⋅10k−410k−19+1210k−19+1=410k−19⋅10k+810k−19+1.

44...488...89 has n+1 numbers "4", n numbers "8", and the "9". So:

44...488...89=4⋅10n−19⋅10n+1+8⋅10n−19⋅10+9

Now, we say 10n=y so

4⋅(10y−1)⋅10y+8⋅10(y−1)+819=400y2+40y+19=(20y+13)2

Note that as y=(10n), 3|(20y+1), for any n value.

I hope it help for you Thanks ☺☺