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# A sequence of 4 digits, when considered as a number in base 10 is four times the number it represents in base 6. What is the sum of the digits of the sequence?(CAT, 2016)

- May 15, 2018
- Posted by: allexamshelps
- Category: CAT

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** Solution:**

**A sequence of 4 digits, when considered as a number in base 10 is four times the :**

** In base n, the four digits are n**^{0}= 1, n^{1}= n, n^{2}= 2n, n^{3}= 3n. That is if you want to write 4 digits in base n, then exponents of n will start from 0 to 3.

^{0}= 1, n

^{1}= n, n

^{2}= 2n, n

^{3}= 3n. That is if you want to write 4 digits in base n, then exponents of n will start from 0 to 3.

**Let's understand it:**

**Further, the question says that there are four digits in a sequence; let’s say the four unknown digits are: pqrs**

**Following the above rule, the four digits in base 6 are:**

**216p+36q+6r+s**

**And, in base 10 are:**

**1000p+100q+10r+s**

**It is given that 10base is 4 times of 6base:**

**That is,**

**10base is 4 x 6base or 10base = 4 (6base); ****or **

**4 x (6base) = 10base **

** (a=b and b=a are equal)**

**4 x (216p+36q+6r+s) = 1000p+100q+10r+s**

**(Open the bracket multiplying each term into the bracket by 4 which is the common multiplier.**

** 4 x (216p+36q+6r+s) = 864p+144q+24r+4s**

**864p+144q+24r+4s=1000p+100q+10r+s**

**Arrange like terms together**

**(864p-1000p)+(144q-100q)+(24r-10r)+(4s-s) = 0**

**-136p+44q+14r+3s=0**

**44q+14r+3s=136p**

**44q+14r+3s=136p now we have to choose values for p, q, r, and s arbitrarily, so that this equation may hold true.**

**By trial and error approach, we have found that p=1, q=2, r=3, and s=2**

**( Actually what we did that we tried to equate 44q+14xr+3s to 136p, and we had put p=1 before. We then found that q=2; r=3 and s=2 can make L.H.S= R.H.S (136)**

**44 x 2+14 x 3+3 x 2=136 x 1**

**88+42+6=136**

**136=136**

**This means p=1; q=2; r=3 and s=2**

**When we add these digits:**

**P+q+r+s=1+2+3+2=8**

**Hence the option (d) is correct.**

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