Let be a very large number which is formed by the concatenation of multiples of 6. What will be the digit of this number when counted from left hand side?
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N = 6 1 2 1 8 2 4 3 0 ⋯
1.
Since N is very large number that is formed by the concatenation of multiples of 6 . Counting all the multiples of 6 is tedious to find the 1 0 0 0 t h digit in the large number N .
In the number above 6 is the one digit number in N . The two digit numbers that are the mutiples of 6 in between 1 0 and 1 0 0 initiate from 6 × 2 = 1 2 and terminates at 6 × 1 6 = 9 6 . Altogether there are 1 6 − 1 = 1 5 (excluding 6 being one digit number) two digits numbers and 1 5 × 2 = 3 0 + 1 = 3 1 digits in the concatenation from 1 to 1 0 0 that are multiple of 6 . Three digits numbers between 1 0 0 and 1 0 0 0 that are mutiples of 6 are 1 6 6 − 1 6 = 1 5 0 (excluding 1 one digit and 15 two digits number). Total digits are 1 5 0 × 3 + 3 1 = 4 8 1 in large number N . To meet the 1 0 0 0 t h digit in N , remaining digits are 1 0 0 0 − 4 8 1 = 5 1 9 which are formed by the concatenation of four digits numbers. Here, 5 1 9 ≡ 3 m o d ( 4 ) where 1 2 9 are pairs of four digits numbers. These 1 2 9 pairs of four digit numbers lies in between 1 6 7 and 3 0 0 . Since 5 1 9 (remaining pairs of 4 digits) is not divisible by 4 however, 5 2 0 = 1 0 0 1 − 4 8 1 or 5 1 6 = 9 9 7 − 4 8 1 is divisible by 4 which implies that 9 9 7 t h and 1 0 0 1 t h digit will be 2 9 5 × 6 = 1 7 7 0 and 2 9 6 × 6 = 1 7 7 6 respectively. So 1 0 0 0 t h digit left hand side of N = 6 1 2 1 8 2 4 3 0 ⋯ 1 7 7 0 1 7 7 6 will be 7 .
2.
Total one digit number that is multiple of 6 be w . Then, w = ⌊ 6 1 0 ⌋ = 1 × 1 = 1 Total two digits numbers x = ⌊ 6 1 0 0 ⌋ − 1 = 1 6 − 1 = 1 5 ⇒ 2 x = 2 × 1 5 = 3 0
Total three digits numbers y = ⌊ 6 1 0 0 0 ⌋ − 1 6 = 1 5 0 ⇒ 3 y = 4 5 0
Concatenation of one, two and three digits that are multiple of 6 are 1 + 3 0 + 4 5 0 = 4 8 1 .
Since there are huge numbers of four digits between 1 0 0 0 and 1 0 0 0 0 digit numbers so such a mutiple of 6 should be obtaines very close to 1 0 0 0 t h and to meet 1 0 0 0 t h digit in the above concatenation required numbers of are 1 0 0 0 − 4 8 1 = 5 1 9 that total four digits numbers ie 4 z = 5 1 9 ⇒ z = ⌊ 4 5 1 9 ⌋ = 1 2 9 pairs of four digits numbers that start from 1 6 7 = 1 6 7 × 6 = 1 0 0 2 to 2 9 6 × 6 = 1 7 7 6 that represents a pair of 4 from 9 9 7 t h to 1 0 0 1 t h digits in concatenation of multiples of 6 .
Therefore, 1 0 0 0 t h digits is 7 .