How many integer solutions exists for the equation $latex \left| y-\sqrt{(y+2)^{2}-9} \right|=5$?

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$latex \left| y-\sqrt{(y+2)^{2}-9} \right|=5$ can be written as:

$latex y-\sqrt{(y+2)^{2}-9}=\pm 5$

$latex \therefore -\sqrt{(y+2)^{2}-9}=\pm 5-y$

$latex \therefore \sqrt{(y+2)^{2}-9}=(y\pm 5)$

Squaring both sides, we get:

$latex \therefore (y+2)^{2}-9=(y\pm 5)^{2}$

$latex \therefore (y+2)^{2}-9=(y\pm 5)^{2}$

$latex \therefore (y+2)^{2}-9=(y\pm 10y+25)$

$latex \therefore (y+2)^{2}=(y\pm 10y+34)$

$latex \therefore y^{2}+4y+4=y^{2}+34\pm 10y$

$latex \therefore \pm 10y+4y=30$

$latex \therefore 14y=30$ or $latex 6y=30$

$latex \therefore y=\frac{30}{14}=\frac{15}{7}$ or $latex y=\frac{30}{6}=5$

Although the equation gives two answers, but only 5 is an integer, there is only one integer solution for the equation.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 100 seconds.*

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For $latex \left| y-\sqrt{(y+2)^{2}-9} \right|=5$, $latex \sqrt{(y+2)^{2}-9}$ should be a perfect square of = 0.

Equating $latex \sqrt{(y+2)^{2}-9}$ to 0, we get:

$latex \sqrt{(y+2)^{2}-9}$ = 0

$latex (y+2)^{2}-9$ = 0

$latex \therefore (y+2)^{2}=9$

$latex \therefore (y+2)=\pm 3$

$latex \therefore y=$ 1or –5

Substituting *y* = 1 in the original equation, we get:

$latex \left| 1-\sqrt{0} \right|=5$, but this cannot be true, as $latex 1\ne 5$, hence, 1 is not the correct solution to the equation.

Substituting *y* = –5 in the original equation, we get:

$latex \left| -5-\sqrt{0} \right|=5$, this is true, as the equation is satisfied. Hence, –5 is the correct solution to the equation. Thus there is only a single solution to the equation.

Note that we are not checking by equating with perfect squares, as:

$latex \sqrt{(y+2)^{2}-9}$ = 4 (a perfect square)

$latex (y+2)^{2}-9$ = 16

$latex \therefore (y+2)^{2}=25$

$latex \therefore (y+2)=\pm 5$

$latex \therefore y=$ 3 or –7

Substituting *y* = 3 in the original equation, we get:

$latex \left| 3-\sqrt{4} \right|=1$, but this cannot be true, as $latex 1\ne 5$, hence, 3 is not the correct solution to the equation.

Substituting *y* = –7 in the original equation, we get:

$latex \left| -7-\sqrt{4} \right|=9$, but this cannot be true, as $latex 9\ne 5$, hence, –7 is not the correct solution to the equation.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 15 seconds.*

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For multiplying any number with 12.

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** Download Practice sheet for MULTIPLYING WITH 12**

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- There is no real need to add the zero if you can remember to perform same steps for all the digits of the number (including the first and the last).
- Make sure to take care of the carryovers as you add the neighboring digits.

Multiplying with 11: (Mul) Nx11

Multiplying with 13: (Mul) Nx13

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For dividing any 2-digit number by 9.

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** Download Practice sheet for DIVIDING 2-DIGIT NUMBER BY 9**

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- When the sum of the 2 digits is ≥ 9, further divide the sum using this method until you get the sum < 9. Add each Quotient you get along the way to get the final Quotient.
- If the sum of the 2 digits is = 9, add 1 to the Quotient and the Remainder becomes 0.

None

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]]>For multiplying any number with 999.

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** Download Practice sheet for MULTIPLYING WITH 999**

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- The separator should be inserted before the hundred’s (100’s) digit.
- The complement of a number is what you need to add to it to make it up to some specified total – in this case 1000.

Multiplying with 9: (Mul) Nx9

Multiplying with 99: (Mul) Nx99

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]]>How many three digit numbers are there in which all the digits are even??

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The answer is 100. Here’s how…

The first digit (hundred’s digit) can have the numbers 2, 4, 6, and 8 – so **4 unique numbers**

The second digit (ten’s digit) can have the numbers 0, 2, 4, 6, and 8 – so **5 unique numbers**

The third digit (one’s digit) can have the numbers 0, 2, 4, 6, and 8 – so **5 unique numbers**

This makes 4 x 5 x 5 = 100 such numbers!

]]>What is the largest number you can make using numbers 5, 1 , 9 and 2 only once? (Psst…. it’s not that easy… there is a catch)

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Most people would see this question and reply 9521 as the largest number, but we already gave you the hint – that it’s not that easy… there is a catch.

So here’s what the real solution is: [latex]((2)^5)^9)^1[/latex]

Do you know the value of this number?

Here it is – **35,184,372,088,832**

In other words – **35 trillion, 184 billion, 372 million, 88 thousand and 832**

Now, can anyone else make a number bigger than this using the same set of digits? The challenge is still on….

Oh, and can you tell the smallest whole number that can be made using the same digits? Just for fun….!

]]>A team played 10 games and scored 1, 2 3, 4 , 5, 6, 7 8, 9 and 10 points. They lost by 1 point in exactly 5 games. In each of the other games they scored two times the points of its opponent. What is the total score of the opponent in all 10 games?

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Let’s call the team whose score is given as Team A and the opposing team whose score we need to find as Team B.

From the question, we already know that in 5 (of the 10) games Team A scored exactly twice that of Team B. Since only even numbers are multiples of 2, these 5 games are those where team A scored even points. In that case, Team B would have scored half of the points Team A scored i.e. 1/2 of 2, 4, 6, 8 and 10.

So the total points of Team B for these 5 games = 1+2+3+4+5 = 15 points

For the remaining 5 games, Team B scored exactly 1 point more than Team A. So, if Team A scored 1, 3, 5, 7 and 9 points, Team B correspondingly scored 1+1, 3+1, 5+1, 7+1 and 10+1 points.

So the total points of Team B for these 5 games = 2+4+6+8+10 = 30 points

The total points scored by Team B will be 15 + 30 = **45 points**

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]]>What comes next in the sequence?

0

10

1110

3110

132110

13123110

??

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The first number is “0”, so we say there is “one 0”, so we get “10” as the next set.

Now we have 10, which is “one 1 and one 0”, so we get “1110” as the next set.

Now we have 1110, which is “three 1’s and one 0”, so we get “3110” as the next set.

Now we have 3110, which is “one 3, two 1’s and one 0”, so we get “132110” as the next set.

Now we have 132110, which is “one 3, one 2, three 1’s and one 0”, so the answer is we get “**13123110**” as the next set.

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