Question

A three digit number is equal to 17 times the sum of its digits; If the digits are

reversed, the new number is 198 more than the old number ; also the sum of

extreme digits is less than the middle digit by unity. Find the original number​

Answer 1

QUESTION -

A three digit number is equal to 17 times the sum of its digits; If the digits are reversed, the new number is 198 more than the old number ; also the sum of extreme digits is less than the middle digit by unity. Find the original number.

ANSWER -

100a+10b+c=17a+17b+17c

100a+10b+c=17a+17b+17c100a+10b+c+198=100c+10b+a

100a+10b+c=17a+17b+17c100a+10b+c+198=100c+10b+aa+c=b-1

I will look at equation 2, removing 10b from both sides

100a+c+198=100c+a

99a+198–99c=0

99(a+2-c)=0

a+c-2=0

a+2=c

Now I will take that information and apply it to equation 3

a+a+2=b-1

b=2a+3

100a+10(2a+3)+(a+2)=17a+17(2a+3)+17(a+2)

I will distribute

100a+20a+a+30+2=17a+34a+17a+51+34

And combine like terms

121a+32=68a+85

53a=53

a=1

SO, let’s solve for b

2a+3=b

2+3=b

b=5

SO, let’s solve for c

a+2=c

1+2=c

c=3

SO, the number is 153

(I could be totally wrong, really)

(Let’s see!)

1+5+3=9

9*17=153

The first statement holds true

198+153=351

The second statement holds true

1+3=4=5–1

The third statement holds true

The 3 digit number is 153

Answer 2

Answer:

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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.

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