The product of the units digit, the tens digit, and the hundreds digit

Question

The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?

(1) m is odd.
(2) The hundreds digit of m is 8

walker · Accepted Answer

I'll try:

1. Let m=htu: h*t*u=96
2. 96=2*2*2*2*2*3

3.1. consider first condition: m is odd.
3.2. u can be 1 or 3, because if m were 5, for example, we would see 5 as a prime number in the product.
3.3. consider u=1. h*t=96 but it is greater than the max possible product: 9*9=81.
3.4. u=3 remains. sufficient

4.1. consider second condition: h=8
4.2. h*t*u=2*2*2*2*2*3
4.3. we have t*u=2*2*3. Both t=2, u=6 and t=4, u=3 satisfy second condition but m have different units digits.
4.4. Insufficient.

5. Therefore, Answer A.

Hope this help

dineshnaidu2410 · Answer

The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?

(1) m is odd.
(2) The hundreds digit of m is 8

Question inference-
Let m= abc ( where a,b,c are the units in hundreds, tens and units place respectively)
Now, from question we know that axbxc = 96

also, a,b,c, are digits and hence belongs to the set { 0,1,2,3,4,5,6,7,8,9 }

Factorizing 96 we get 96= 2^5 x 3^1
Only possible solution sets of abc from above equations are { 2, 8, 6} , { 4, 4, 6}, { 8, 4, 3 }

Since, 2 x 8 x 6, 4 x 4 x 6, 8 x 4 x3 = 96

Now statement 1 : 
 
(1) M is odd, hence units digit is odd and only solution set where one number is odd is { 8, 4, 3 }hence unit digit = 3 ( Sufficient)

Now statement 2 :
(2) Hundred digit of m is 8,

we have 2 solution sets where 8 appears => { 2,8,6} & { 8,4,3}

hence it is clearly insufficient

Correct Answer is A

CaspAreaGuy · Answer

it is C.
the number is 843.

statement1, although useful still  insuff icient. the only odd factor in 96 is 3. so if units digit is 3, the product of the first two ones must be 32 (32*3=96).

statement2 says hundreds is 8. alone is  insuff

but both statements are  suff . 843

marcodonzelli · Answer

OA is A

if m is odd, then the unit digit is 3, since 96= 2^5 *3. any other combination would give us an even number

Ravshonbek · Answer

1. prime factors m= x  y  z 
 96-2
48-2
24-2
12-2 
 6-2 
 3-3 
1-1

so: hundreds and tens digits can be either 8 or 2 then units digit has to be 3.

suff.

2. m=xyz then y=8, then x and z can either be 4 or 3 so we cannot say anything from this statement. insuff.

A for me.

miketal20 · Answer

Agree with Marco. A.

2^5X3, You can't have 2 (or any of its multiples e.g. 4, 8, etc) as Units digit, otherwise it will be even.

icaniwill · Answer

96 = 2^5 * 3

1) m is odd
 so units digit can be 1 or 3 based on the factors. but it cannot be 1 as no two single digits integers can give a product of 96.
So units digit is 3 - SUFFICIENT

2) hundreds digit is 8
Insufficient as number could be 843 or 826 or 834 or 862

Ans: A

icaniwill · Answer

96 = 2^5 * 3

1) m is odd
so units digit can be 1 or 3 based on the factors. but it cannot be 1 as no two single digits integers can give a product of 96.
So units digit is 3 - SUFFICIENT

2) hundreds digit is 8
Insufficient as number could be 843 or 826 or 834 or 862

Ans: A

Ben10512 · Answer

Dear Experts,

In my opinion the answer should be (c).

The factors are 2^5*3 and I found 5 cases.

286,446,826,483,843.

As per A) there are two cases that end with odd number 483 & 843.

As per B) there are two cases that begin with 8 those are 826 and 843.

Choice A + B - we have 843. Satisfies both condition.

But the answer is A.

Posted from my mobile device

Bunuel · Answer

What is the  units digit  of m?

In both cases the units digit is 3.

Ben10512 · Answer

I understand that we need to find a number that has units digit as 3 but if you observe there are two cases with option A. Hence the answer cannot be A.

Posted from my mobile device

Bunuel · Answer

The question asks to find the units digit of m, not m itself. The units digit of m in both cases is 3.

MikeCorleone · Answer

Let m = cba = 96

(1) a = 1,3,5,7,9 --> 5,7,9 cannot divide 96. Thus 1,3 remians
if a= 1, then ab = 96 (not possible because maximum we can get is 9x9 =81)
thus a = 3 Answer --> Sufficient

(2) Given b= 8, then ca = (4,3) , (3,4), (6,2), (2,6) --> Not Sufficient

kawal27 · Answer

STATEMENT 1:  SUFFICIENT 
prime factorization of 96 =2*2*2*2*2*3
  now you can see only 8,6,4,2 and 3 can place at units digit. 
 
 from which only 3 is odd

STATEMENT 2: INSUFFICIENT 
8*6*2
8*2*6
8*4*3
8*3*4

there are many possibilities  
ANS :A

ManishaPrepMinds · Answer

The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?

(1) m is odd.
(2) The hundreds digit of m is 8

From the question stem: 
1) m=xyz
2) x*y*z=96=2*2*2*2*2*3

Question: z=?

Analyze statement (1) ALONE 
(1) m is odd.
 Possible values of z are (1,3,5,7,9) 
But (5,7,9) are not possible as there is no 5, 7 and one more extra 3 to get 9 in the above Prime Factorization of 96
=> Thus Possible values of z are 1 and 3 
 Now if z =1  then x*y=96=2*2*2*2*2*3 which yield the combinations such as (32,3), (16,6), (12,8) etc which is not possible as x and y are non zero digits
 => thus z=3  then x*y= 32 which gives only possible combinations for (x,y)= (8,4) or (4,8)
Thus answer to our question is z=3
 Hence statement (1) ALONE is sufficient

Analyze statement (2) ALONE 
(2) The hundreds digit of m is 8
=> y*z=2*2*3
=>  Possible Combinations of (y,z)= (2,6) or (4,3)
=> z= 6 or 3 
As we do not get the unique value of z
 Hence statement (2) ALONE is NOT sufficient

ANSWER: A

CEdward · Answer

Here is how I thought about it.

The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?

h x t x u = 96 = even
1. even x even x even = even
2. odd x even x even = even (or some rearrangement of this)

(1) m is odd.

This implies that u = 1,3,5,7, or 9

96 = 8 x 4 x 3

Therefore, u MUST be 3. It can't be 1 because the only way to get 96 then would be 8 x 12 (but 12 is clearly 2 digits). It can't be 5,7, or 9 either b/c there are no two numbers that can be multiplied with them to get 96.

(2) The hundreds digit of m is 8

8 4 3 
8 3 4

Insufficient.

A

deeuce · Answer

Hi  walker  thanks for the answer. Would you kindly help me understand the max possible product part 9*9 = 81 ? I must be missing something with my evening fried brain.

Thanks

bambinobirbante · Answer

Please help in clarification -

Consider m = abc

Therefore, a * b * c = 96

Statement A : m is odd

factors of 96 = 1 *  2 * 2 * 2 * 2 * 2  * 3
Since m is odd, c can only be 1 or 3 because multiplier of 2 will make it even
If c = 1, a * b = 96 ==> (a,b) or (b,a) could be  (1,96), (2,48), (3,32), (4,24), (6,16), (8,12)  
Since a and b are digits therefore they can not be more than 9. Hence, c = 3

SUFFICIENT

Statement B : a = 8 ==> b * c = 12
Now (b,c) or (c,b) could be ( 1,12), (2,6),   (3,4) 
Using statement 2 --> b = 4 and c = 3 satisfy the condition. Therefore, Statements A & B together also are SUFFICIENT.

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The product of the units digit, the tens digit, and the hundreds digit

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