Each of the integers from 0 to 9, inclusive, is written on a separate

Hi Bunnel,
This solution means we are actually able to decide what we are going to pick. Right?
I some how got the idea we can not see which slip we are going to pick, so my answer was 9.
If we cannot decide which one to pick then will 9 be correct?

Nope. The question asks "how many must be drawn to ensure that..."

To ensure that the sum will be 10 we should consider the worst possible scenario.

Similar questions to practice:
12-easy-pieces-or-not-126366.html#p1033935
in-his-pocket-a-boy-has-3-red-marbles-4-blue-marbles-and-85216.html
of-the-science-books-in-a-certain-supply-room-50-are-on-131100.html
in-a-deck-of-52-cards-each-card-is-one-of-4-different-color-83183.html
a-box-contains-10-red-pills-5-blue-pills-12-yellow-56779.html
each-of-the-integers-from-0-to-9-inclusive-is-written-on-130562.html
a-student-is-asked-to-pick-marbles-from-a-bag-that-contains-72390.html
of-the-science-books-in-a-certain-supply-room-50-are-on-131100.html
if-a-librarian-randomly-removes-science-books-from-a-library-93861.html
m10-q24-ps-69233.html#p1237169

Hope it helps.

You can make a sum of 10 by pairing these numbers:
1, 9
2, 8
3, 7
4, 6

So if you pick only one number from each of these pairs and the leftover 2 numbers: 0 and 5, you would have picked 6 numbers without any two numbers adding up to 10. When you pick the next number (7th), there will be one of the 'sum 10' pairs. So picking 7 numbers will give you one of the required pairs for sure!

Answer (E)

Just a view :

Even after drawing 7 if we pair it with 0 we will not get a 10 , Drawing 7 does not ensure that sum of ANY of the drawn slips will be 10.
Yes, we can make a 10 but we can also NOT do so.
Isn't the question asking how many we must draw to ensure (100%) that sum of ANY of the 2 drawn slips must be 10?

Maybe I am over reading the word " ANY".

How many must be withdrawn so that we CAN make a pair with a sum 10 seems , more appropriate.

Hello

I think the question does NOT mention the word ANY. The question clearly states, "....how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10". So if we pick 7 slips, we can be completely sure that we will be able to find two slips here which will add up to '10'. And in worst case scenario, 7 slips are what are required to be drawn.

Amazing question | Sharing the method to solve backwards from answer options

Option A - Say we pick up (0, 1, 2) or (3, 4, 5) then we do not get a 10 whereas if we pick up (4, 5, 6) or (0, 1, 9) then we get a sum of 10 from two slips

Option B - Say we pick up (0, 1, 2, 3) or (2, 3, 4, 5) then we do not get a 10 whereas if we pick up (3, 4, 5, 6) or (0, 1, 2, 9) then we get a sum of 10 from two slips

Option C - Say we pick up (0, 1, 2, 3, 4) or (1, 2, 3, 4, 5) then we do not get a 10 whereas if we pick up (2, 3, 4, 5, 6) or (0, 1, 2, 3, 9) then we get a sum of 10 from two slips

Option D - Say we pick up (0, 1, 2, 3, 4, 5) then we do not get a 10 whereas if we pick up (1, 2, 3, 4, 5, 6) or (0, 1, 2, 3, 4, 8) then we get a sum of 10 from two slips

Option E - Say we pick up (0, 1, 2, 3, 4, 5, 6) or (1, 2, 3, 4, 5, 6, 7) or (4, 5, 6, 7, 8, 9, 0) or any other combination of 7 integers between 0 and 9, incl. we will get at least 1 pair of integers that sums to 10

Ans. E

We have 10 different slips of paper with the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 listed. How many must we draw to ensure that 2 of the numbers will total 10?

The only way we can arrive at a sum of 10 from 2 slips is if we have one out of {6, 7, 8, 9}.

We have to consider the worse case scenario: if we draw all the slips aside from {6, 7, 8, 9}.

6 + 1 = 7 to ensure that 2 numbers will total 10.

Always think of the worst case in such questions.

The worst case is when he starts from smallest number up, Thus until he reaches 5 (0,1,2,3,4,5), no 2 numbers can make 10
It is only when he picks up any other number, be it 6 or 7 or 8 or 9, he will be able to complete the sum of 2 chits as 10.

This 6+1 = 7

Imagine you having the worse lucky picking by chance the slip of papers. Here is the sequence of the most unlucky person alive:

First drawn: number 0
Second drawn: number 1
Third drawn: number 2
Fourth drawn: number 3
Fifth drawn: number 4
Sixth drawn: number 5
Seventh drawn: number 6

4 + 6 = 10

E

I used this same technique in other similar GMAT official question. GMAT is not about tricks. It’s about patterns.

You can solve it using the pigeonhole method.

1+9, 2+8, 3+7, 4+6. There are 4 possibilities.

Start off with the worst-case scenario: Draw 0 and 5 in the first two tries.
Since there are 4 possibilities, we need 5 draws to ensure that the sum is 9.
So, 2 + 5 = 7.

Option E

­If we pick the numbers 0, 1, 2, 3, 4, and 5, none of these pairs sum to 10.Now, the next (7th) number we draw, whether it is 6, 7, 8, or 9, will ensure that we have a pair that sums to 10:

• If we draw 6, it pairs with 4.
• If we draw 7, it pairs with 3.
• If we draw 8, it pairs with 2.
• If we draw 9, it pairs with 1.

So, indeed, drawing 7 slips will guarantee that at least one pair of numbers sums to 10. Thanks for pointing that out!