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Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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29 Sep 2013, 07:55
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Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed? A. 1 B. 2 C. 3 D. 4 E. 5
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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nave81 wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 Given digits are: {4, 5, 7, 8, 9}. In order the units digit of the sum to be 5, the units digits of the two integers must be 7 and 8. Thus 2 cases are possible: AB7 +C8 ___ 555 OR: AB8 +C7 ___ 555 Clearly A must be 4: 4B7 +C8 ___ 555 OR: 4B8 +C7 ___ 555 B and C can be 5 and 9 or viseversa. Therefore there are total of 4 pairs: (457, 98), (497, 58), (458, 97), and (498, 57). Answer: D.
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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03 May 2015, 19:19
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because we have 555 the unit digit must be 7 and 8 the ten digit must be 14 because on is transfered. so they are 5 and 9 the hundred must be 4 so 4 (5,9) and (8, 7)
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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13 Sep 2015, 16:42
nave wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 Let a,b,c,d, and e represent a distinct digit from the above set. abc+ de555 Then 100*a+10*(b+d)+c+e=555The terms 100*a and 10*(b+d) must end with zero, because they are multiples of 100 and 10 respectively. So it follows that c+e must end with a unit digit of 5. The smallest c+e can be is 9(5+4) and the biggest it can be is 17(8+9). The only number with a units digit of 5 between 9 and 17 is 15. From the given set, it follows that either c=7 or 8 and e=8 or 7. Then 100*a+10*(b+d)=540 and the remaining digits that can be used are 5,9, and 4. If a=5 then b+d should be equal to 4, but this is not possible from the remaining digits to choose from. a must be equal to 4. Therefore b=5 or 9 and d=5 or 9. There are 4 numbers that satisfy the above equation. Correct answer is D. Similar question to practice: http://www.artofproblemsolving.com/wiki ... Problem_15



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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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13 Sep 2015, 22:37
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nave wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 Another method would be to start from the hundreds digit. A three digit number and a two digit number add to give 555. So the three digit number must be greater than 455 but less than 500 since you do not have smaller digits to make one of the early 500 numbers such as 510. So the hundreds place of three digit number must have 4. You are left with 5, 7, 8, 9 Next, the units digits must add to 15 and there is only one way of doing that: 7+8 = 15. The two digits, 7 and 8 can be arranged in either way  4 __ 7 or 4 __ 8 so there are two ways to choose the units digit. The leftover digits are 9 and 5 which need to go to tens places. Again, there are two ways of arranging them 4 9 __ or 4 5 __. In all, there are 2*2 = 4 ways of making a 3 digit and a 2 digit number such that the sum is 555. Answer (D)
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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21 May 2017, 19:08
nave wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 1. Since the sum is 555 and ends with 5, the two numbers that form the unit digits in the 3 digit and 2 digit number must be 8 and 7. 2, We can see that first digit of the three digit number has to be 4. 3. We can easily list the numbers as 498 and 57, 458 and 97, 497 and 58, 457 and 98
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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05 Sep 2017, 08:23
nave wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 Hi.. 1) Hundreds digit cannot be anything other than 4.. 1 way2) Units digit can be ONLY 8 and 7, one in 3digit number and other in 2digit.... 2ways3) Tens digit  the remaining two can also be any one in 3digit and other in 2digit.. 2 ways ans 1*2*2=4 D example  138+27 = 137+28= 128+37=127+38
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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19 Oct 2017, 06:01
Bunuel wrote: nave81 wrote: Each of the digits 7, 5, 8, 9 and 4 is used only one to form a three digit integer and a two digit integer. If the sum of the integers is 555, how many such pairs of integers can be formed?
A. 1 B. 2 C. 3 D. 4 E. 5 Given digits are: {4, 5, 7, 8, 9}. In order the units digit of the sum to be 5, the units digits of the two integers must be 7 and 8. Thus 2 cases are possible: AB7 +C8 ___ 555 OR: AB8 +C7 ___ 555 Clearly A must be 4: 4B7 +C8 ___ 555 OR: 4B8 +C7 ___ 555 B and C can be 5 and 9 or viseversa. Therefore there are total of 4 pairs: (457, 98), (497, 58), (458, 97), and (498, 57). Answer: D. The posts by Bunuel are so crisp and methodical ! Thanks for being on the forum.



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Re: Each of the digits 7, 5, 8, 9 and 4 is used only one to form [#permalink]
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