Question

you guys never help me ....... plz this time plz help me.... but i know again no-one is going to help me let's see


The question is in attachment​

Answer 1

Let the number be abc

So number is 100a+10b+c

⇒100a+10b+c=17(a+b+c)

⇒83a=7b+16c .......(i)

198+100a+10b+c=a+10b+100c

Or 198+99a=99c

Hence c−a=2⇒c=a+2 .....(ii)

Also, given a+c=b−1

Now as a+c=b−1 and c=a+2

⇒a+a+2=b−1

⇒b=2a+3

now in eq(i)

substitute the values of b and c

83a=7b+16c

⇒83a=7(2a+3)+16(a+2)

⇒53a=21+32

⇒a=1

⇒c=a+2=3

⇒b=2a+3=5

So the number is 153

Answer 2

Answer:

the answer of this question is 135