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In a three digit positive integer, the tens digit is the average of th

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Bunuel
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In a three digit positive integer, the tens digit is the average of th [#permalink] New post   01 Jun 2020, 10:41
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In a three digit positive integer, the tens digit is the average of the other two digits. The ratio of the number formed by its first two digits to their sum equals the ratio of the number formed by its last two digits to their sum. How many three digit numbers satisfy these conditions?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10


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D
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   01 Jun 2020, 11:58
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IMO D

Let ABC be 3 no. Digit
100A+10B+C= 3 no. Digit.
B= A+C/2 --------(1)


Also, The ratio of the number formed by its first two digits to their sum equals the ratio of the number formed by its last two digits to their sum.
Such that 10A+B/A+B = 10B+C/B+C
10A+B(B+C) = 10B+C(A+B)
10AB+10AC+B^2+BC=10AB+AC+10B^2+BC
9AC=9B^2 => AC = B^2 -----(2)
=> AC = (A+C/2)^2 (from equation 1)
=> 4AC = A^2 + C^2 + 2AC
=> 0 = A^2 + C^2 - 2AC
=> 0 = (A-C)^2 => A=C------(3)
Putting value in equation 2
=> A^2 = B^2 => A=B --------(4)
=> from equation 3&4
=> A=B=C
So the no. Satisfying all such conditions= 111,222,333,444,555,666,777,888,999 = 9 no.
=> Answer is D

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In a three digit positive integer, the tens digit is the average of th [#permalink] New post   01 Jun 2020, 12:32
Picking up numbers and solving can be quite easy.
Let the digit be : XYZ, where x is hundred's digit, y is ten's and z is one's place digit

Given Y is average of x and z : It means X and Z can be even numbers only. Cannot be odd or different odds, except in case of triplets which satisfy this condition of Y as average of X and Z
Let check numbers except triplets
2 Y 4 : Number will be 234
4 Y 6 :Number will be 456
6 Y 8 :Number will be 678
and reversing hundred and ones place: we have 6 digits :

Now lets check 2nd condition of
xy/ x+y = yz / y+z
23/5 = 34/7
67/13 = 78/ 15

Similarly all 6 numbers don't satisfy this condition,
You can come to this conclusion once you test 2-3 possible values, No need to check all and waste time.

Now lets check triplets.
2 Y 2 : Number will be 222
3 Y 3 :: Number will be 333
4 Y 4: Number will be 444
.
.
9 Y 9 :: Number will be 999
So, we have 9 numbers satisfying Condition 1, Lets check condition 2
22/4 = 22/4
33/6= 33/6
and so on will be same for all 9 numbers

Answer D : 9
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   03 Jun 2020, 09:32
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IMO D

Let the three digit number be abc
a = hundreds digit
b = Tens digit
c = Units digit

Not Given : b= (a+c)/2. ---- eq 1

Also ab/ (a+b) = bc / (b+c)

it can also be written as (b+c)/bc = (a+b)/ab

or: 1/c + 1/b = 1/b + 1/a
or: 1/c = 1/a
=> a=c -- eq 2

Now from eq 1 & eq2 ==> a=c=b

So all three digits of the number are same: {111,222,333,444,555,666,777,888,999} : Total 9
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   09 Jun 2020, 02:57
NitishJain wrote:
IMO D

Let the three digit number be abc
a = hundreds digit
b = Tens digit
c = Units digit

Not Given : b= (a+c)/2. ---- eq 1

Also ab/ (a+b) = bc / (b+c)

it can also be written as (b+c)/bc = (a+b)/ab

or: 1/c + 1/b = 1/b + 1/a
or: 1/c = 1/a
=> a=c -- eq 2

Now from eq 1 & eq2 ==> a=c=b

So all three digits of the number are same: {111,222,333,444,555,666,777,888,999} : Total 9


Hi NitishJain,

The highlighted statement will be \(\frac{10a+b}{a+b} = \frac{10b+c}{b+c}\)
From above equation, we can get \(b^2=ac\) ----(1)
Again, we know \(2b=a+c\) ----(2)

From 1 and 2, we get a=c=b. So, all 3 digits same. 9 such numbers.

Although the eventual result obtained is same in this scenario, it maybe give erroneous result in some other scenario. Agree?

Thanks
Lipun
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   09 Jun 2020, 05:36
Expert Reply
Bunuel wrote:
In a three digit positive integer, the tens digit is the average of the other two digits. The ratio of the number formed by its first two digits to their sum equals the ratio of the number formed by its last two digits to their sum. How many three digit numbers satisfy these conditions?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10


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The tens digit is the average of the other two digits.
Implication:
The three digits are all the same, or the middle digit is halfway between the other two digits.
111, 123, 135, 147, etc.

The ratio of the number formed by the first two digits to the sum of the first two digits equals the ratio of the number formed by the last two digits to the sum of the last two digits.
In the list above, only the green integer satisfies this condition, implying that all three digits must be the SAME:
111, 222, 333, 444, 555, 666, 777, 888, 999

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D
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NitishJain
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   09 Jun 2020, 11:16
Lipun wrote:
NitishJain wrote:
IMO D

Let the three digit number be abc
a = hundreds digit
b = Tens digit
c = Units digit

Not Given : b= (a+c)/2. ---- eq 1

Also ab/ (a+b) = bc / (b+c)

it can also be written as (b+c)/bc = (a+b)/ab

or: 1/c + 1/b = 1/b + 1/a
or: 1/c = 1/a
=> a=c -- eq 2

Now from eq 1 & eq2 ==> a=c=b

So all three digits of the number are same: {111,222,333,444,555,666,777,888,999} : Total 9


Hi NitishJain,

The highlighted statement will be \(\frac{10a+b}{a+b} = \frac{10b+c}{b+c}\)
From above equation, we can get \(b^2=ac\) ----(1)
Again, we know \(2b=a+c\) ----(2)

From 1 and 2, we get a=c=b. So, all 3 digits same. 9 such numbers.

Although the eventual result obtained is same in this scenario, it maybe give erroneous result in some other scenario. Agree?

Thanks
Lipun


Hello Lipun,

you might be correct.

GMATWhizTeam: Could you please advise here.
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   11 Jun 2020, 07:03
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NitishJain wrote:
Lipun wrote:
NitishJain wrote:
IMO D

Let the three digit number be abc
a = hundreds digit
b = Tens digit
c = Units digit

Not Given : b= (a+c)/2. ---- eq 1

Also ab/ (a+b) = bc / (b+c)

it can also be written as (b+c)/bc = (a+b)/ab

or: 1/c + 1/b = 1/b + 1/a
or: 1/c = 1/a
=> a=c -- eq 2

Now from eq 1 & eq2 ==> a=c=b

So all three digits of the number are same: {111,222,333,444,555,666,777,888,999} : Total 9


Hi NitishJain,

The highlighted statement will be \(\frac{10a+b}{a+b} = \frac{10b+c}{b+c}\)
From above equation, we can get \(b^2=ac\) ----(1)
Again, we know \(2b=a+c\) ----(2)

From 1 and 2, we get a=c=b. So, all 3 digits same. 9 such numbers.

Although the eventual result obtained is same in this scenario, it maybe give erroneous result in some other scenario. Agree?

Thanks
Lipun


Hello Lipun,

you might be correct.

GMATWhizTeam: Could you please advise here.


Hi NitishJain,
Your approach was right till \(b = \frac{a+c}{2} ….Eq.(i)\)
However, when you are writing a two digit integer, like “ab” or “bc”, you have to keep in mind that “ac” is not the product of a and c and therefore, you cannot treat a and c in “ac” as individual numbers. Let’s understand it with example. Suppose abc = 333.
    • So, when you are saying, \(\frac{(b+c)}{bc} = \frac{(a+b)}{ab}\)
      o It means, \(\frac{(3+3)}{33} = \frac{(3+3)}{33}\) it does not mean \(\frac{(3+3)}{3*3 }= \frac{(3+3)}{3*3} = \frac{3}{3*3 }+ \frac{3}{3*3} ⟹\frac{1}{3 }+ \frac{1}{3}\)
      o And thus, carrying out the operation in following manner is incorrect: \(\frac{(3+3)}{33} = \frac{3}{33 }+ \frac{3}{33} ⟹\frac{1}{3 }+ \frac{1}{3}\)
        o the correct result is \( \frac{3}{33 }+ \frac{3}{33} =\frac{1}{11 }+ \frac{1}{11}\)
    • So, to avoid the confusion, we write two digit number “ac” with the help of place value as \(10*a +c\)
      o In the above example we can write 33 as \(3*10 +3\), which is equal to 33.
I hope it helps! :)

Regards,
GMATWhiz team
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   03 Jul 2020, 04:06
Can I have similar questions like this I am facing difficulty to solve this kind of questions Bunuel

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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   03 Jul 2020, 04:17
Let 3 digit of the number is abc
given:
1. b=a+c/2
2. {ab/(a+b)}={bc/(b+c)}
how many Three digit number satisfies above 2 equation??
Cross multiplying equation 2 for further solving
ab^2+abc=abc+b^2c
a=c

it means a,b,c digits are same. So from 1 to 9 = Nine cases are possible (111,222,333,-----, 999)

Answer 9 (D)
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink] New post   23 Mar 2024, 16:39
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Re: In a three digit positive integer, the tens digit is the average of th [#permalink]
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