Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Grade 10 Cayley #25 The digits of the positive integer n include no 9s, exactly four 8s, exactly three 7s, exactly two 6s, and some other digits. If the sum of the digits of n is 104 and the sum of the digits of 2n is 100, then the number of times that the digit 5 occurs in n is...?

Meta-math
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

for the sake of argument lets assume that all the digits of n from 1-> 9 be separated by 0's. i.e. 1020304050606 We know that there are 4 8's, 3 7's 2 6's \[104 = 8(4) + 7(3) + 6(2) +X_{1}\]\[X_{1} = 39\]Where X1 is the sum of the digits less that 6. Notice that \[8 * 2 = 16\]\[1 + 6 = 7\] \[7 * 2 = 14\]\[1 + 4 = 5\] \[6 * 2 = 12\]\[1 + 2 = 3\] \[5 * 2 = 10\]\[1 + 0 = 1\] Since we put a 0 between the digits the double of the number will not "overflow" to the other digits. using or example earlier. Twice of it s 2040608101212. So we could say, \[100 = 7(4) + 5(3) + 3(2) + X_{2}\]\[X_{2} = 51\] Note that lets say i have a number A wherein all its digits are less that 5. The sum of digits of 2A is equal to the double of the sums of digits of A because\[4 * 2 = 8\]\[3 * 2 = 6\]\[2 * 2 = 4\]\[1 * 2 = 2\] So we could say that \[X_{1} = a + 5b\]\[X_{2} = 2a + b\] Where a is the sum of all the digits less than 5 and b is the number of 5's in n. Solving for b \[39 = a + 5b\]\[51 = 2a + b\] \[3 = b\] So the answer is there are 3 5's in n.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question