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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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30 Oct 2017, 07:08
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The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers? A. 21 B. 25 C. 41 D. 67 E. 73
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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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30 Oct 2017, 07:29
swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67
Can someone please provide me the solution to this question. the answer depends on properties of number... first there are five odd digits... 1,3,5,7,9.. SUm of which 4 consecutive gives you multiple of 10.... it is 7+9+1+3=20.. so the numbers will be TWo with one lower TENS digit  X7 and X9 and two with one higher TENS digit  (X+1)1 and (X+1)3 so BASICALLY we are adding SUM of tens digit {x+x+(x+1)+(x+1)}*10 SuM of units digit is 7+9+1+3=20... so when you add this to TENS sum, we get \((x+x+(x+1)+(x+1))*10 + 20 = 10x+10x+10(x+1)+10(x+1) +20=10x+10+10x+10+10(x+1)+10(x+1) = 10(x+1)+10(x+1)+10(x+1)+10(x+1)=40(x+1)\) so \(\frac{40*(x+1)}{10} = 4*(x+1)\) should be PERFECT SQUARE.. so x+1 can be 4 as 4*4 will be perfect square.. so numbers are 37,39,41,43 ans C
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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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21 Jul 2018, 19:07
swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67 E. 73 let sequence=x, x+2, x+4, x+6 sum of sequence=4x+12 if 4x+12 is a multiple of 10, then x must have a units digit of 7 4*17+12=80; 80/10=8 no 4*27+12=120; 120/10=12 no 4*37+12=160; 160/10=16 yes sequence=37, 39, 41, 43 41 C




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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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19 Jul 2018, 19:34
I tried the following method which is similar to chetan2u 's solution but maybe takes a little longer. Let's say the 4 consecutive odd integers are 2k+1, 2k+3, 2k+5 & 2k+7. It says [(2k+1)+(2k+3)+(2k+5)+(2k+7)]/10 = a^2x*b^2x which is in the form of a perfect square. > (8k+16)/10=a^2x*b^2y > 8(K+2)/10=a^2x*b^2y > 4(K+2)/5=a^2x*b^2y The first number at which 4(k+2)/5 is a perfect square is when (K+2)/5=4. This means K=18 So our numbers are 2(18)+1, 2(18)+3, 2(18)+5 & 2(18)+7 = 37,39,41 & 43. Don't forget the kudos if this helped you.



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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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Updated on: 24 Jul 2018, 08:00
swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67 E. 73 pushpitkc, Bunuelmy reasoning was following say the first three consecutive twodigit odd integers are 11, 13, 15 so if you sum up these numbers we get 39 now i need to try answer choices to find the 4th odd number ...OA is C so if add up 39+41=80 > 80/10 = 8 and that is not a perfect square. i think something wrong is with this question hello generis, my last hope why everyone starts counting one digit numbers ? do you have your unique constructivelly structured approach to this question ? i switched to number properties questions and i think i dont have a good "number sense" though yes i know divisibility rules, i know what is factor and what is multiple ... Factors are what we can multiply to get the number. Multiples are what we get after multiplying the number by an integer etc Shine on me crazy diamond i like begining of the long version of this song
Originally posted by dave13 on 22 Jul 2018, 04:14.
Last edited by dave13 on 24 Jul 2018, 08:00, edited 1 time in total.



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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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22 Jul 2018, 08:32
rajudantuluri wrote: I tried the following method which is similar to chetan2u 's solution but maybe takes a little longer. Let's say the 4 consecutive odd integers are 2k+1, 2k+3, 2k+5 & 2k+7. It says [(2k+1)+(2k+3)+(2k+5)+(2k+7)]/10 = a^2x*b^2x which is in the form of a perfect square. > (8k+16)/10=a^2x*b^2y > 8(K+2)/10=a^2x*b^2y > 4(K+2)/5=a^2x*b^2y The first number at which 4(k+2)/5 is a perfect square is when (K+2)/5=4. This means K=18 So our numbers are 2(18)+1, 2(18)+3, 2(18)+5 & 2(18)+7 = 37,39,41 & 43. Don't forget the kudos if this helped you. This is a very well thought out approach. Can solve it in a minute .It took me 4 minutes with mine, I got lost in between.



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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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23 Jul 2018, 22:34
gracie wrote: swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67 E. 73 let sequence=x, x+2, x+4, x+6 sum of sequence=4x+12 if 4x+12 is a multiple of 10, then x must have a units digit of 74*17+12=80; 80/10=8 no 4*27+12=120; 120/10=12 no 4*37+12=160; 160/10=16 yes sequence=37, 39, 41, 43 41 C Could you please explain the highlighted part?



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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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23 Jul 2018, 23:36
Swatipupi wrote: gracie wrote: swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67 E. 73 let sequence=x, x+2, x+4, x+6 sum of sequence=4x+12 if 4x+12 is a multiple of 10, then x must have a units digit of 74*17+12=80; 80/10=8 no 4*27+12=120; 120/10=12 no 4*37+12=160; 160/10=16 yes sequence=37, 39, 41, 43 41 C Could you please explain the highlighted part? hi Swatipupi, in order for 4x+12 to be a multiple of 10, then 4x must have a units digit of 8 because x is odd, only a units digit of 7 will produce this I hope this helps, gracie



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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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23 Jul 2018, 23:52
Thank you Gracie!! Silly me!
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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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24 Jul 2018, 00:27
Multiply each option choice by 4 to reach close to the sum of the consequtive numbers, than using approximation choose what number upon diving by 10 could possibly be a perfect square. Evaluate the answer choice & you're good to go.
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The sum of 4 consecutive twodigit odd integers, when divided by 10,
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29 Aug 2018, 01:18
swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67 E. 73 I solved it by taking a range of min and max for the sum of 4 twodigit consecutive odd nos, i.e (11+13+15+17 = 10 * 4 + 16 = 56) and (99+97+95+93 = 400  16 = 384). Dividing the upper and lower range by 10, the perfect square value should be between 5.6 and 38.4. Means, 9,16,25 and 36. From the sum of four no, the mean of 4 no. can be found which would be 22.5, 40, 62.5 and 90. As the four no.s are consecutive odds, the mean would be even, i.e 40 and 90. So, the no. would be any from these 37,39,41,43,87,89, 91 and 93. From the options, only (C) 41 satisfy. Interestingly, while solving I read the no. as consecutive rather than consecutive odd. Hence, marked 21 as for regular consecutive no.s the avg would be either 22.5 or 62.5 and only 21 satisfies it.



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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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30 Jan 2019, 04:31
Solution:Let the four consecutive two digit odd integers be x; x+2; x+4; x+6. \(Sum = 4x + 12.\) Given: when 4x + 10 is divided by 10; becomes perfect square. Here we can do this problem by unit digit concept; 4x + 10 must be divided by 10 means the unit digit of “x” must be 7. Let’s substitute and check the sequence which will fit the above condition: i) \(4x+12=4 ×17+12= \frac{80}{10} =8\);Not perfect square ii) \(4x+12=4 ×27+12= \frac{120}{10} =12\);Not perfect square iii) \(4x+12=4 ×37+12= \frac{160}{10} =16\);Perfect square Hence the required sequence of two digit consecutive odd integers = 37; 39; 41; 43. So, “41” can possibly be one of these 4 integers. The correct answer option is “C”.
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Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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30 Jan 2019, 15:34
chetan2u wrote: swatygi wrote: The sum of 4 consecutive twodigit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?
A. 21 B. 25 C. 41 D. 67
Can someone please provide me the solution to this question. the answer depends on properties of number... first there are five odd digits... 1,3,5,7,9.. SUm of which 4 consecutive gives you multiple of 10.... it is 7+9+1+3=20.. so the numbers will be TWo with one lower TENS digit  X7 and X9 and two with one higher TENS digit  (X+1)1 and (X+1)3 so BASICALLY we are adding SUM of tens digit {x+x+(x+1)+(x+1)}*10 SuM of units digit is 7+9+1+3=20... so when you add this to TENS sum, we get \((x+x+(x+1)+(x+1))*10 + 20 = 10x+10x+10(x+1)+10(x+1) +20=10x+10+10x+10+10(x+1)+10(x+1) = 10(x+1)+10(x+1)+10(x+1)+10(x+1)=40(x+1)\) so \(\frac{40*(x+1)}{10} = 4*(x+1)\) should be PERFECT SQUARE.. so x+1 can be 4 as 4*4 will be perfect square.. so numbers are 37,39,41,43 ans C Hello chetan2u !!! My approach was almost the same but I got lost in the end. (n2) + (n) + (n+2) + (n+4) = 4(n+1)... This must be div by 10. From here I just picked up letter C cuz it was a four, obviously, i was wrong (I was running out of time). Could you please explain to the following? "so x+1 can be 4 as 4*4 will be perfect square.. so numbers are 37,39,41,43"Kind regards!




Re: The sum of 4 consecutive twodigit odd integers, when divided by 10,
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