Turito Academy

Test Prep

Global form fields

The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K) Where 0  xi 9, i = 1, 2, …6, where 0 &lt; K &lt; 18 = Coefficient of xK in (1 + x + x2 +….. + x9)6 = Coefficient of xK in <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAFQAAAAsCAYAAAD7Gp9tAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAABGJhU0UAAAAeOiuv/QAAAvJJREFUeNrtWz1IHEEUHo7DQo4DEZEQ0qQIIiLp5AgigRQSrrUIwUIEkRRikSZIELFJlSKIjYQgIjYhhCA2IsH+kMNCRBCxSCEHIiISRLi8h6/Ym9vJztvbn5nd+eCDvd27d3PfvXnv7bxZIRzCYgS4D/wLbBKVeAdcybFYX0gDFQaBB8AhHWOjwDqwmGNB8bcfAscU19dJp0B0A8+Azy0V4hnwIzmEH6rAY2I1wNYQaVHyudYAvgWeA0+AEyojy8BViz1rAzijiGcVinm9xN90LmjqL/ucvweuAcvALjqekt/UB7wCPs7AlPUT9Kc0hXHK/gqw8wh4Tdp4gV5Z8Lwukbe2YIH+YZFRQa8lEQp0LghfKYx4gR7Z43mNXlqTP3ikG2g145Vpgvqdu9Ow9YJirhfDlJjKlMCwIhqXxWlEGK9s8NCipociLoAD0jlMSqd0rS0pYUDdjHDwNsRQPN7WtLfpl3T+B3TfWUMFnQZ+UlQk04wxVSiz91CSwYz/mjGGdc6gtxnG0/BQvCvpl2bUimIsMuU6FDP0JXCO8f0YH3c4A274lAYmCYo/6DMdvwLuJRxCStwccysF7KgFbWowCLsU9+pUmCeJAmmUiJcllZTmqcwZTinR3WVJUKwFv9O0f2NQ5WCloE/pFrGbYlkthSmfGUH7KLv2Spl6I4uChkksHOD98Q/FYs2WfKuXJQ/NC5ygTlAnqBPUCWqIoM2c0Hmom/JOUAcnqBM0e4LexzAAk9vMYcBaD+1kxV4Fk9vMXBQFc8X+j2jdCZHaVDEUuHzI6ilh13PcUkHDtJm5wI4wq+v5TTAb+YZ5qG6bOSxmSCNtTIr4VsGz0Gbe4jocZuSLFAU1vc2M8XOA+yHcAl2x1EMRcbWZX4r23XdaeC8e9kLaKGicbWYMhR/ClgZx7GC2uc38BHjTib1FEf3jNDa3mXGn8lInBvBfPo0oDtneZsYnYVRPgbDjUd6fU+qiJD0alUEsZFdzLCiGvVAbkP8BBkUbZzCPY0kAAAD1dEVYdE1hdGhNTAA8bWF0aCB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMTk5OC9NYXRoL01hdGhNTCI+PG1zdXA+PG1mZW5jZWQgc2VwYXJhdG9ycz0ifCI+PG1mcmFjPjxtcm93Pjxtbj4xPC9tbj48bW8+LTwvbW8+PG1zdXA+PG1pPng8L21pPjxtbj4xMDwvbW4+PC9tc3VwPjwvbXJvdz48bXJvdz48bW4+MTwvbW4+PG1vPi08L21vPjxtaT54PC9taT48L21yb3c+PC9tZnJhYz48L21mZW5jZWQ+PG1uPjY8L21uPjwvbXN1cD48L21hdGg+JwI/oAAAAABJRU5ErkJggg==" class="Wirisformula" alt="open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent" width="84" height="44" style="vertical-align:-14px" role="math" data-mathml='«math xmlns=¨http://www.w3.org/1998/Math/MathML¨ »«msup»«mrow»«mfenced separators=¨|¨»«mrow»«mfrac»«mrow»«mn»1«/mn»«mo»-«/mo»«msup»«mrow»«mi»x«/mi»«/mrow»«mrow»«mn»10«/mn»«/mrow»«/msup»«/mrow»«mrow»«mn»1«/mn»«mo»-«/mo»«mi»x«/mi»«/mrow»«/mfrac»«/mrow»«/mfenced»«/mrow»«mrow»«mn»6«/mn»«/mrow»«/msup»«/math»'/> = Coefficient of xk in (1 – 6x10 + 15 x20 – ….) (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …) = k + 6CK – 6. K–4CK–10 = k + 6C6 – 6. K–4C6 .

The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 &lt; K &lt; 18) is -

Solution for the question - the numbers of integers between 1 and 106 have the sum of their digit equalto k(where 0 < k < 18) is - k + 6c6 6 k 4c6