What are the last two digits (at ten's place and unit’s place) in the

Question

What are the last two digits (at ten's place and unit’s place) in the numerical value of the following product?

243 x 5787 x 5315 x 7779 x 8262 x 3179 x 993.

A. 10
B. 30
C. 50
D. 70
E. 90

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EgmatQuantExpert · Accepted Answer

Given

• 243 x 5787 x 5315 x 7779 x 8262 x 3179 x 993.

To Find

• The last two digits of the number.

Approach and Working Out

• A very interesting question which looks overwhelming at the first go

• One approach could be to manipulate and multiply the numbers with unit digits as 5 (5315) and 2 (8262) first.

o 243 × 5787 × 5315 × 7779 × 8262 × 3179 × 993.
o = 243 × 5787 × 7779 × 3179 × 993 × (5315 × 8262)
o = 243 × 5787 × 7779 × 3179 × 993 × (XYZ … 30)

• The last digit of the above expression will be 0.

o The second last digit will be the units digit of,
o = 243 × 5787 × 7779 × 3179 × 993 × XYZ … 3
o = 3 × 7 × 9 × 9 × 3 ×3 
o = 9

• Combining we get the last two digits as 90.

• However, this manipulation can be done as we have the luxury of using two numbers to get a 0.

Correct Answer: Option E

Fido10 · Answer

Man ! what is this explanation !!

lacktutor · Answer

243 x 5787 x 5315 x 7779 x 8262 x 3179 x 993=

—> (..3*...7*...9*...9*...3)* (...15*...62)

= (...3 )* (...93*10) = ...9*10 = ...90

The answer is E.

Posted from my mobile device

exc4libur · Answer

Hey, thanks for posting.

Could you explain why you multiplied the units of some numbers and the tens–units of others?

catinabox · Answer

Can someone help with this question?

I'm getting 30 as the last 2 digits; not sure how the 9 came about.

priyamitra · Answer

I did the longer version as that is the only method that came to mind. 
Its safer to use parenthesis or you may get really confused

Taking the last two digits of each number

(43 x 87) x (15 x 79) x (62 x 79) x 93

= 37 41  x 11 85  x 48 98  x  93

-> (41 x 85) x (98 x 93)

= 34 85  x 91 14

-> 85 x 14 = 11  90

Ans - E

Do note that u dont need to completely multiply the numbers. Once you have the numbers upto 10's place; you can move to the next step

GMATGuruNY · Answer

\frac{243 * 5787 * 5315 * 7779 * 8262 * 3179 * 993}{100}   =   \frac{243 * 5787 * 1063 * 7779 * 4131 * 3179 * 993}{10}

Rule:
When a positive integer is divided by 10, the remainder is equal to the units digit.

Units digit for  243*5787*1063*7779*4131*3179*993  = units digit for 3*7*3*9*1*9*3 --> 3*7*9*9*3*3 = 21*81*9 --> units digit of 9
Thus:
  \frac{243 * 5787 * 1063 * 7779 * 4131 * 3179 * 993}{10}   --> remainder of 9

Rule:
If  \frac{x}{y}  yields a remainder of R, then  \frac{kx}{ky}  yields a remainder equal to  kR .
For example:
 \frac{10}{7}  --> remainder of 3
 \frac{20}{14}  --> remainder of 6
 \frac{30}{21}  --> remainder of 9
In other words:
If the numerator and denominator each increase by a factor of k, the remainder also increases by a factor of k.

The blue fraction above yields a remainder of 9.
In the original green fraction, the numerator and denominator are each multiplied by a factor of 10.
Thus, the remainder for the green fraction must be 10 times the remainder for the blue fraction:
10*9 = 90

E

Fdambro294 · Answer

fantastic approach! kudos! Picked up so much useful information from just this 1 Post.

drjob0105 · Answer

I would like to know why was the equation divided by 100 in the first step! Can someone please explain

Posted from my mobile device

Fdambro294 · Answer

I would like to know why was the equation divided by 100 in the first step! Can someone please explain

Posted from my mobile device   [/quote

Hey,

I can try to help. What he is using is 2 rules

1st, he’s using the Divisibility and Remainder Rule when you divide by 10. Whenever you divide by 10, the units digit will be the remainder. This makes it easy to just find the units digit of the product and call that the remainder when dividing by 10.

2nd, he uses the rule that says if you multiply the NUM and DEN by 10, the remainder you get will ALSO be multiplied by 10

The reason why he starts out with 100 is because the Divisibility and Remainder Rule for 100 similarly says that the last 2 Digits will be the remainder when you divide by 100.

9,125 / 100. —— Remainder = 25 (the last 2 digits)

To find the last 2 digits by dividing that entire long product by 100 would take time. Effectively that is what most of us did. Took the last 2 digits of every number and multiplied them all to find the final 2 digits of the result.

Instead, 1st, he started out with 100 in the denominator. Then he divided the original long Product in the NUM by 5 and 2 (look at the multiple of 5 and the even number on the right vs left)

Similarly he divided the DEN of 100 by 5 and 2 to get the DEN down to 10.

2nd, he then finds the units digit of the long Product in the Factored down NUM. It is 9. This is the Remainder when the Factored Down NUM is divided by 10

3rd, which is the Rule I never really thought about, if you multiply the NUM back up by a factor of 10 and multiply the DEN of 10 by 10, the other Rule says that the Remainder is also multiplied by 10

Now we have the full ORIGINAL multiplication chain divided by 100 with a remainder = 9 * 10 = 90.

Using that Divisibility and Remainder Rule of 100 again, the last 2 Digits of the entire Multiplication Chain when divided by 100 will give you the remainder.

Thus the last 2 digits are 90

Didn’t realize until I started how involved the explanation would be. I hope that helped somewhat?

GMATGuruNY · Answer

Fdambro294 has done a great job explaining the logic.
When an integer with at least two digits is divided by 100, the remainder is equal to the last two digits of the integer:
 \frac{137}{100}  --> remainder = 37
 \frac{8512}{100}  --> remainder = 12
 \frac{9273481}{100}  --> remainder = 81
Implication:
We can obtain the last two digits of the given product by determining the remainder when the product is divided by 100.

Kinshook · Answer

Asked: What are the last two digits (at ten's place and unit’s place) in the numerical value of the following product?

243 x 5787 x 5315 x 7779 x 8262 x 3179 x 993 = (243 x 5787 x 7779 x 3179 x 993) x (5315 x 8262) 
Digit of unit's place = 0
Digit of tenth place = unit digit of (3*7*9*9*3*3) = 9

IMO E

HamP95 · Answer

If you have excellent mental maths, this one is easily doable by multiplying the first 2 equations by the last 2 digits and "carrying over" the answer to the last two digits of the next number, as an example:

43 * 87 = 3741

Next is 5315, so take the 41 from 3741 and multiply by 15 to get 615

Then take the 79 from 779 and multiply by 15 to get 1185

From there, it goes

85 * 62 = 5270
70 * 79 = 5530
30 * 93 = 2690

last two digits = 90

Note this method will require excellent mental maths to complete in the time required on GMAT, but if you're confident in your head multiplication this made it very easy for me.

Ilanchezhiyan · Answer

how do you deduce if the numbers are not divisible by 10, what if its not possible to write the numbers as divisible by 100 and by 10

Bunuel · Answer

Experts Global Video Explanation:

https://www.youtube.com/watch?v=zJ09xBoGcqo

What are the last two digits (at ten's place and unit’s place) in the

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