Digits Flashcards
If the two digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N?
A) 181 B) 165 C) 121 D) 99 E) 44
Let the first digit be xy, then the second digit would be yx.
xy = 10x + y yx = 10y + x
sum = 11 (x+y)
so the sum has to be multiple of 11. A is not.
1,234 1,243 1,324 ..... .... \+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exactly once in each integer. What is the sum of these 24 integers?
A. 24,000 B. 26,664 C. 40,440 D. 60,000 E. 66,660
Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.
The sum is therefore = (1 + 10 + 100 + 1000) * (16 +26 +36 +46) = 1111 * 6 * 10 = 66660
Answer : e
In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and multiplies the integer on the card by the next larger integer. If each possible product is between 15 and 200, then the least and greatest integers on the cards could be
A. 3 and 15 B. 3 and 20 C. 4 and 13 D. 4 and 14 E. 5 and 14
Given: 15 discard A and B;
If x=4 then x(x+1)=20>15 –> so, the least value is 4, discard E. Test for the largest value: if x=14 then x(x+1)=14*15=210>200 –> discard D.
Answer: C.
Else you could find that the greatest value is 13 and since only C offers it, then it must be correct.
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704 B. 2,990 C. 5,404 D. 5,444 E. 5,994
Since every digit has equal likely chances of appearing at any of the three places in a 3 digit number
so out of 27 number
9 will have Unit digit 1, 9 will have unit digit 2 and 9 will have unit digit 3
So sum of the unit digits of all 27 numbers = 9(1+2+3) = 54
We write 4 at the unit digit place and 5 goes carry forward
so out of 27 number
9 will have Tens digit 1, 9 will have Tens digit 2 and 9 will have Tens digit 3
So sum of the Tens digits of all 27 numbers = 9(1+2+3) = 54
but 5 is carry forward so the sum becomes 54+5 = 59, We write 9 at the tens digit place and 5 goes carry forward
so far the sum looks like _ _ 9 4
This much calculation gives us the answer out of given options however we can do the same thing two more times to find exact sum which will be 5,994
Or you can simply recognize that 27=8+9+10, so x=8-1=7.
Answer: option E
Which of the following is the product of two integers whose sum is 11?
(A) -42 (B) -28 (C) 12 (D) 26 (E) 32
a+b=11
a*b=42
a*(11-a) = 42
a^2 - 11a + 42 = 0
(a-14)(a+3) –> a=14 is a valid solution. therefore, b=-3 …. a+b = 14-3 = 11 and ab=(14)(-3) = -42
AB
+BA
———–
AAC
In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?
A. 9 B. 6 C. 3 D. 2 E. 0
The sum of 2 two-digit integers cannot be greater than 200 (actually it cannot be greater than 99+99=198).
Thus from: AB \+BA \_\_\_ AAC
It follows that A=1. So, we have that 1B \+B1 \_\_\_ 11C
If B is less than 9, AB+BA will not be a three-digit integer (consider 18+81=99), thus B=9 –> 19+91=110 –> C=0.
Answer: E.
If the product of the integers w, x, y and z is 770, and if 1 < w < x < y < z, what is the value of w + z?
A. 10 B. 13 C. 16 D. 18 E. 21
770 = 2 * 5 * 7 * 11
So w = 2, x = 5, y = 7, z = 11
w + z = 2 + 11 = 13
Answer – B
3P5
+4QR
——–
8S4
In the correctly worked addition problem, P, Q , R and S are digits. If Q= 2P what is the value of S?
Units digit: 5 + R results in 4 => The only possible scenario is that R = 9 and 5+R = 14, of which 1 is a carry in the tens position
For the tens place: 1 + P + Q = S (with or w/o a carry)
However, in the hundreds place: 3 + 4 gives 8 => there was a carry of 1
=> In the tens place: 1 + P + Q = S + 10
=> P + Q = S + 9
=> P + 2P = S + 9
=> 3P = S + 9
=> S must be a multiple of 3
=>S = 0 or 3 or 6 or 9
Answer A
A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player’s current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player’s current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game
i. 0.8
ii. 0.5
iii. 0.1
A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii
the Q may look very lengthy and full of values..
But here we can eaily find a solution if we understand the concepts..
three scenarios..
1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1
2) if the number rolled times 0.1 is less than the player’s current number and is odd, the players number is decremented bu 0.1
3) if the number rolled times 0.1 is less than the player’s current number and is even, the players number is unaffected…
then there are some %s given…
What can we finally deduct.. THE MAX VALUE... YES, the max value can be 0.7... reason:- after a person has got 0.6, there is only one value, which can increase 0.6 and that is by throwing 6.. this will take the value to 0.7 but cannot increase it further as there is no 7 on dice.. so 0.8 is out.. only D is without (i) ans D
Each of the nine digits 0, 1, 1, 4, 5, 6, 8, 8, and 9 is used once to form 3 three-digit integers. What is the greatest possible sum of the 3 integers?
A. 1,752 B. 2,616 C. 2,652 D. 2,775 E. 2,958
Since we need to maximize the sum, first we will maximize the sum of hundreds digits of the 3 numbers. Hundred digits of 3 numbers will be 8, 8 and 9 in any order.
Now we will try to maximize the sum of tens digits of 3 numbers. Tens digits of 3 numbers will be 4, 5 and 6 in any order.
Remaining digits can be put in units place in any order.
Sum of 3 numbers= 100(8+8+9) + 10(4+5+6) + (0+1+1) = 2500+150+2 = 2652
AB
x BA
____
The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?
(A) 11 (B) 12 (C) 13 (D) 21 (E) 31
First of all, since the digits must be distinct, then we can eliminate option A (11).
Next, simply plug in the options and see which satisfies AB x BA = ACA:
(B) 12 –> 1221 –> the units digit is not 1. Discard.
(C) 13 –> 1331 –> the units digit is not 1. Discard.
(D) 21 –> 21*12 = 252 = AB x BA = ACA. Bingo.
Answer: D.
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
Given that (10a + b) - (10b + a) = 27 –> 9a - 9b =27 –> a - b = 3.
Answer: A.
If the tens digit x and the units digit y of a positive integer n are reversed, the resulting integer is 9 more than n. What is y in terms of x?
A. 10 - x B. 9 - x C. x + 9 D. x - 1 E. x + 1
n = 10x + y and n’ = 10y + x
n’ - n = (10y + x) - (10x +y) = 9 –> x + 1
E.
If x = 1 / (2^2 * 3^2 * 4^2 * 5^2) is expressed as a decimal, how many nonzero digits will x have ?
Terminating Decimal–> REF FRACTIONS AMC
ANS: C
N and M are each 3-digit integers. Each of the numbers 1, 2, 3, 6, 7, and 8 is a digit of either N or M. What is the smallest possible positive difference between N and M?
A. 29 B. 49 C. 58 D. 113 E. 131
You have 6 digits: 1, 2, 3, 6, 7, 8
Each digit needs to be used to make two 3 digit numbers. This means that we will use each of the digits only once and in only one of the numbers. The numbers need to be as close to each other as possible. The numbers cannot be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other.
The first digit (hundreds digit) of both numbers should be consecutive integers i.e. the difference between 1** and 2** can be made much less than the difference between 1** and 3. This gives us lots of options e.g. (1 and 2) or (2 and 3) or (6 and 7) or (7 and 8**).
Now let’s think about the next digit (the tens digit). To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible (1 is possible) and the tens digit of the smaller number should be as large as possible (8 if possible). So let’s not use 1 and 8 in the hundreds places and reserve them for the tens places since we have lots of other options (which are equivalent) for the hundreds places. Now what are the options?
Try and make a pair with (2** and 3). Make the 2 number as large as possible and make the 3** number as small as possible. We get 287 and 316 (difference is 29) or
Try and make a pair with (6** and 7). Make the 6 number as large as possible and make the 7** number as small as possible. We get 683 and 712 (difference is 29)
The smallest of the given options is 29 so we need to think no more. Answer must be (A).
The question is not a hit and trial question. It is completely based on logic and hence do not ignore it.
