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# What are the last two digits of (301*402*503*604*646*547*448*349)^2

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Math Expert
Joined: 02 Sep 2009
Posts: 50712
What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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07 Apr 2015, 05:12
3
10
00:00

Difficulty:

95% (hard)

Question Stats:

39% (02:15) correct 61% (02:06) wrong based on 224 sessions

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What are the last two digits of $$(301*402*503*604*646*547*448*349)^2$$

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

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Joined: 02 Sep 2009
Posts: 50712
Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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13 Apr 2015, 06:29
3
4
Bunuel wrote:
What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.

Find remainder of (301*402*503*604*646*547*448*349)^2/100

Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.

We need to find the remainder of:

(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100

This implies we need to find the remainder of:

(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50

We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.

We need the remainder of:

(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50

Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:

We need:

the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50

the remainder of (12)*(24)*(24)*(24) / 50

Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.

We need:

the remainder of (6)*(24)*(24)*(24) / 25

the remainder of (6)*(25-1)*(25-1)*(25-1) / 25

Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:
the remainder of 6*(-1)*(-1)*(-1) / 25

The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76

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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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07 Apr 2015, 06:17
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2
What are the last two digits of (301*402*503*604*646*547*448*349)^2

One thing to note before I start: if you got 2 numbers with unknown digits except for last 2, you can then determine the last 2 digits of their product. For example, xxxxxx36 * xxxx47 = xxxxxxxx92 regardless of "x" where x - digits.
Lets change our number a little - as in, cut away useless information and leave only tenth and unit digits
$$(301*402*503*604*646*547*448*349)^2$$ -> $$(1*2*3*4*46*47*48*49)^2$$ -> $$(24*46*47*48*49)^2$$ -> $$(4*47*48*49)^2$$ -> $$(88*48*49)^2$$ -> $$(24*49)^2$$ -> $$76^2$$ -> $$76$$- our last 2 digits
B that is
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Joined: 07 Aug 2011
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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07 Apr 2015, 07:28
1
Bunuel wrote:
What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

The last two digits of a product of two numbers are dependent on the last 2 digits of the individual numbers. so .
301*349 = XXXX49
402*448 = XXXX96
503*547 = XXXX41
604*646 = XXXX84

now the question is what are the last 2 digits of 49*96*41*84 for that i have used a non-trivial method which i learnt in school days . attached the snapshot for rest of solution .
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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07 Apr 2015, 12:11
1
2
((301*402*503*604*646)*(547*448*349))^2

If you observe above digits, last digit are: 1,2,3,4,6,7,8,9; 5 is missing; so I have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01,02,03,04,46 and final three as 47*48*49.

Solving for only last two digits and multiplying them we get:( (06*04*46)(56*49))^2 = (44*44)^2=36^2=76

Thanks,
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Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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04 Nov 2016, 05:16
Bunuel wrote:
What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

Slightly different approach.

In order to find out last 2 digit of a product we need to operate only with last two digits.
So we have:

$$\frac{(01*02*03*04*46*47*48*49)^2}{100}$$

Which is same as

$$\frac{1^2*2^2*3^2*4^2*46^2*47^2*49^2}{2^2*25}$$

Now we simplify this fraction cancelling 4 (keeping in mind that we need to multiply by it at the end)
Also noticing that 46 ≡ -4 (mod 25), 47 ≡ -3 (mod 25) and so on.

As a result we have:

$$\frac{1*3^2*4^2*(-4)^2*(-3)^2*(-2)^2*(-1)^2}{25}$$

$$\frac{(9*16*16*9*4)}{25}$$ = (keeping track only of last two digits) = $$\frac{81*56*4}{25}$$ = $$\frac{44}{25}$$

Now we need to multiply back by the cancelled factor $$2^2$$:

$$\frac{44*4}{100} = \frac{76}{100}$$

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Joined: 18 Jun 2017
Posts: 60
Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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02 Aug 2017, 05:10
1
(301*402*503*604*646*547*448*349)^2
Since the last two digits of a product of two numbers are dependent on the last 2 digits of the individual numbers.
So discard each term's hundred's digit and keep multiplying the digits using the ten's and unit's place derived out of each progressing product as we move left to right.
(01*02*03*04*46*47*48*49)^2
01*02*03*04*=24
24*46=0[ten's digit from the product]4[unit's digit from the product]
Likewise 04*47=88
88*48=24
24*49=76
So finally (76)^2 = 76.
Option B.
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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06 Aug 2017, 01:45
Bunuel wrote:
What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

(1*2*3*4*6*7*8*9)^2
=(144*56*9)^2
=xx576^2 (P.s.- xx24^odd=xx24 ; xx24^even=xx76)
=76

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Location: India
Concentration: Social Entrepreneurship, General Management
GMAT 1: 690 Q49 V34
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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10 Aug 2017, 09:33
Bunuel wrote:
Bunuel wrote:
What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96
(B) 76
(C) 56
(D) 36
(E) 16

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.

Find remainder of (301*402*503*604*646*547*448*349)^2/100

Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.

We need to find the remainder of:

(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100

This implies we need to find the remainder of:

(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50

We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.

We need the remainder of:

(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50

Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:

We need:

the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50

the remainder of (12)*(24)*(24)*(24) / 50

Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.

We need:

the remainder of (6)*(24)*(24)*(24) / 25

the remainder of (6)*(25-1)*(25-1)*(25-1) / 25

Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:
the remainder of 6*(-1)*(-1)*(-1) / 25

The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76

is there a simpler solution ? this is too long n prone to errors .
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2  [#permalink]

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06 Sep 2018, 10:12
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Re: What are the last two digits of (301*402*503*604*646*547*448*349)^2 &nbs [#permalink] 06 Sep 2018, 10:12
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